# Antiderivatives and Indefinite Integration, including Trig Integration

This section covers:

WARNING: The techniques in this section only work if the argument of what’s being integrated is just “$$x$$”; in other words, “$$x$$” is by itself and doesn’t have a coefficient or perhaps more complicated. See the U-Substitution Integration section for more integrating more complicated expressions!

# Antiderivatives (Anti-derivatives)

Antiderivatives are just the opposite of derivatives; it’s that simple. For example, since we know that the derivative of $$5x$$ is 5, the antiderivative of 5 is $$5x$$. But we have to be careful here, since the derivative of $$5x+4$$ is also 5 (since the derivative of a constant is 0). So, technically, the antiderivative of 0 is “$$C$$”, where $$C$$ is any constant, because the derivative of $$C$$ is 0.

When we take the antiderivative of something, we are actually integrating it, or taking the integral of it. The integrand is the name of what we’re taking the integral of.

Here are definitions of an antiderivative, integral, constant of integration, and differential equation. We’ll talk more about Differential Equations in the Differential Equations and Slope Fields section.

Definitions of Antiderivative (Integral):

A function $$F$$ is an antiderivative of another function $$f$$ when $${F}’\left( x \right)=f\left( x \right)$$.

Note that the term indefinite integral is another word for an antiderivative, and is denoted by the integral sign $$\int{.}$$

When we have the differential equation (an equation that involve $$x$$, $$y$$ and the derivative of $$y$$) in the form $$\displaystyle \frac{{dy}}{{dx}}=f\left( x \right)$$, we can write it as $$dy=f\left( x \right)dx$$. When we integrate, we have $$\displaystyle y=\int{{f\left( x \right)dx}}=F\left( x \right)+C$$, where $$C$$ is the constant of integration.

But the most important thing to remember is that integration is the opposite of differentiation!

# Basic Integration Rules

I know this seems confusing at this point, but let’s go through some basic integration rules and examples, and then do some problems!

Basic Integration Rules:

 Differentiation Formula Integration Formula Examples $$\displaystyle \frac{d}{{dx}}\left( \text{C} \right)=0$$ ($$C$$ is a constant) $$\int{0}\,\,dx=\text{C}$$ $$\displaystyle \frac{d}{{dx}}\left( \text{6} \right)=0$$ $$\int{0}\,\,dx\text{ can equal 6}$$ $$\displaystyle \frac{d}{{dx}}\left( {kx} \right)=k$$ ($$k$$ is a constant) $$\int{k}\,dx=kx+C$$ $$\displaystyle \frac{d}{{dx}}\left( {3x} \right)=3$$ $$\int{3}\,dx=3x+C$$ $$\displaystyle \frac{d}{{dx}}\left( {k\cdot f\left( x \right)} \right)=k\cdot {f}’\left( x \right)$$ $$\int{{k\cdot f\left( x \right)dx}}=k\int{{f\left( x \right)dx}}$$ $$\displaystyle \frac{d}{{dx}}\left( {4x} \right)=4\cdot \text{derivative of }x=4\cdot 1=4$$ $$\int{{4\cdot 1\,dx}}=4\int{{\left( 1 \right)dx}}=4x$$ $$\displaystyle \frac{d}{{dx}}\left( {f\left( x \right)\,\pm \,g\left( x \right)} \right)$$ $$\displaystyle \,\,\,\,\,\,\,\,\,\,\,\,\,\,={f}’\left( x \right)\,\,\pm \,\,{g}’\left( x \right)$$ $$\begin{array}{l}\int{{\left( {f\left( x \right)\,\,\pm \,\,g\left( x \right)} \right)}}\,\,dx=\\\,\,\,\,\,\,\,\,\int{{f\left( x \right)dx\pm \,\,\int{{g\left( x \right)}}}}\,dx\end{array}$$ $$\displaystyle \frac{d}{{dx}}\left( {4x+3x} \right)={f}’\left( {4x} \right)\,+\,{g}’\left( {3x} \right)=4+3=7$$ $$\int{{\left( {4+3} \right)}}dx=\int{{4\,dx+\int{3}}}dx=4x+3x$$ $$\displaystyle \frac{d}{{dx}}\left( {{{x}^{n}}} \right)=n{{x}^{{n-1}}}$$ $$\displaystyle \int{{{{x}^{n}}dx}}=\frac{{{{x}^{{n+1}}}}}{{n+1}};\,\,n\ne -1$$ $$\displaystyle \frac{d}{{dx}}\left( {{{x}^{3}}} \right)=3{{x}^{{3-1}}}=3{{x}^{2}}$$ $$\require {cancel} \displaystyle \int{{3{{x}^{2}}dx}}=3\int{{{{x}^{2}}dx}}=3\cdot \frac{{{{x}^{{2+1}}}}}{{2+1}};=\cancel{3}\cdot \frac{{{{x}^{3}}}}{{\cancel{3}}}={{x}^{3}}$$

# Trigonometric Integration Rules

Here are the trigonometry integration rules:

 Trig Differentiation Formula Trig Integration Formula $$\displaystyle \frac{d}{{dx}}\left( {\sin x} \right)=\cos x$$ $$\int{{\cos x}}dx=\sin x+C$$ $$\displaystyle \frac{d}{{dx}}\left( {\cos x} \right)=-\sin x$$ $$\int{{\sin x}}dx=-\cos x+C$$ $$\displaystyle \frac{d}{{dx}}\left( {\tan x} \right)={{\sec }^{2}}x$$ $$\int{{{{{\sec }}^{2}}\,x}}dx=\tan x+C$$ $$\displaystyle \frac{d}{{dx}}\left( {\sec x} \right)=\sec x\tan x$$ $$\int{{\sec x\tan x}}dx=\sec x+C$$ $$\displaystyle \frac{d}{{dx}}\left( {\cot x} \right)=-{{\csc }^{2}}x$$ $$\int{{{{{\csc }}^{2}}x}}dx=-\cot x+C$$ $$\displaystyle \frac{d}{{dx}}\left( {\csc x} \right)=-\csc x\cot x$$ $$\int{{\csc x\cot x}}dx=-\csc x+C$$

Here are some hints to help you remember the trig differentiation and integration rules:

When the trig functions start with “c”, the differentiation or integration is negative (cos and csc). For the functions other than sin and cos, there’s always either one tan and two secants, or one cot and two cosecants on either side of the formula. Look at the formulas and see how this makes sense!

Note: More trig integration rules (involving “ln”) will be introduced later here in the Exponential and Logarithmic Integration section.

# Indefinite Integration Problems

Let’s do some problems; notice how we may need to rewrite the integral, and then simplify at the end. Note to check your work, you can differentiate back from the answer to see if you get the original!

Also, remember that you can check these by going to www.wolframalpha.com (or use app on smartphone) and type in “integral of x^.25” for example.

 Original Integral Rewritten Integral Integrated Answer Simplified Answer $$\int{{\sqrt[4]{x}}}dx$$ $$\int{{{{x}^{{\frac{1}{4}}}}}}dx$$ $$\displaystyle \frac{{{{x}^{{\frac{1}{4}+1}}}}}{{\frac{1}{4}+1}}+C=\frac{{{{x}^{{\frac{5}{4}}}}}}{{\frac{5}{4}}}+C$$ $$\displaystyle \frac{4}{5}{{x}^{{\frac{5}{4}}}}+C$$ $$\displaystyle \int{{\frac{1}{{2{{x}^{5}}}}}}dx$$ $$\displaystyle \frac{1}{2}\int{{{{x}^{{-5}}}}}dx$$ $$\displaystyle \frac{1}{2}\cdot \frac{{{{x}^{{-5+1}}}}}{{-5+1}}+C=\frac{1}{2}\cdot \frac{{{{x}^{{-4}}}}}{{-4}}+C$$ $$\displaystyle -\frac{1}{{8{{x}^{4}}}}+C$$ $$\displaystyle \int{{\frac{5}{{{{{\left( {3x} \right)}}^{3}}}}}}dx$$ $$\displaystyle \int{{\frac{5}{{27{{x}^{3}}}}}}dx=\frac{5}{{27}}\int{{{{x}^{{-3}}}}}dx$$ $$\displaystyle \frac{5}{{27}}\cdot \frac{{{{x}^{{-3+1}}}}}{{-3+1}}+C=\frac{5}{{27}}\cdot \frac{{{{x}^{{-2}}}}}{{-2}}+C$$ $$\displaystyle -\frac{5}{{54{{x}^{2}}}}+C$$ $$\displaystyle \int{{\frac{{2\sqrt[3]{x}}}{{{{x}^{2}}\sqrt{x}}}}}dx$$ $$\displaystyle \begin{array}{c}2\int{{{{x}^{{\frac{1}{3}}}}\cdot {{x}^{{-2}}}\cdot {{x}^{{-\frac{1}{2}}}}dx}}\\=2\int{{{{x}^{{-\frac{{13}}{6}}}}dx}}\end{array}$$ (add exponents when multiplying!) $$\displaystyle 2\cdot \frac{{{{x}^{{-\frac{{13}}{6}+1}}}}}{{-\frac{{13}}{6}+1}}+C=2\cdot \frac{{{{x}^{{-\frac{7}{6}}}}}}{{-\frac{7}{6}}}+C$$ $$\displaystyle -\frac{{12}}{{7{{x}^{{\frac{7}{6}}}}}}+C$$ $$\displaystyle =-\frac{{12}}{{7{{x}^{7}}\sqrt[6]{x}}}+C$$ $$\displaystyle \int{{4\sin x}}dx$$ $$\displaystyle 4\int{{\sin x}}dx$$ $$4\cdot -\cos x+C$$ $$-4\cos x+C$$

Here are more problems:

 Integration Problem Solution $$\int{{\left( {{{x}^{2}}+5x+1} \right)}}dx$$ $$\displaystyle \int{{\left( {{{x}^{2}}+5x+1} \right)}}dx=\int{{{{x}^{2}}dx+5\int{x}dx+\int{1}\,dx}}=\frac{{3{{x}^{2}}}}{3}+\frac{{5{{x}^{2}}}}{2}+x+C$$ Check: $$\displaystyle \frac{d}{{dx}}\left( {\frac{{{{x}^{3}}}}{3}+\frac{{5{{x}^{2}}}}{2}+x+C} \right)=\frac{1}{3}\cdot 3{{x}^{2}}+\frac{1}{2}\cdot 10x+1={{x}^{2}}+5x+1\,\,\surd$$ $$\displaystyle \int{{\left( {\frac{{\sqrt{x}+5}}{{{{x}^{2}}}}} \right)}}dx$$ $$\displaystyle \int{{\left( {\frac{{\sqrt{x}+5}}{{{{x}^{2}}}}} \right)}}dx=\int{{\frac{{{{x}^{{\frac{1}{2}}}}}}{{{{x}^{2}}}}}}dx+\int{{\frac{5}{{{{x}^{2}}}}}}dx=\int{{{{x}^{{-\frac{3}{2}}}}}}dx+5\int{{{{x}^{{-2}}}}}dx$$ $$\displaystyle =\frac{{{{x}^{{-\frac{1}{2}}}}}}{{-\frac{1}{2}}}+\frac{{5{{x}^{{-1}}}}}{{-1}}+C=-\frac{2}{{\sqrt{x}}}-\frac{5}{x}+C=-\frac{{2\sqrt{x}+5}}{x}+C$$ Check: $$\displaystyle \frac{d}{{dx}}\left( {-\frac{{2\sqrt{x}+5}}{x}} \right)=\frac{d}{{dx}}\left( {-2{{x}^{{-\frac{1}{2}}}}-5{{x}^{{-1}}}+C} \right)$$ $$\displaystyle =1\cdot {{x}^{{-\frac{3}{2}}}}-\left( {-5{{x}^{{-2}}}} \right)={{x}^{{-\frac{3}{2}}}}+5{{x}^{{-2}}}$$ $$\displaystyle =\frac{{{{x}^{{\frac{1}{2}}}}}}{{{{x}^{2}}}}+\frac{5}{{{{x}^{2}}}}=\frac{{\sqrt{x}+5}}{{{{x}^{2}}}}\,\,\surd$$ $$\displaystyle \int{{\left( {4{{{\sec }}^{2}}\theta +\cos \theta } \right)}}d\theta$$ $$\displaystyle \int{{\left( {4{{{\sec }}^{2}}\theta +\cos \theta } \right)}}d\theta =4\int{{{{{\sec }}^{2}}\theta }}d\theta +\int{{\cos \theta }}d\theta =4\tan \theta +\sin \theta +C$$ Check: $$\displaystyle \frac{d}{{d\theta }}\left( {4\tan \theta +\sin \theta +C} \right)=4{{\sec }^{2}}\theta +\cos \theta \,\,\surd$$

# Initial Conditions and Particular Solutions

Since when we take an integral of a function, we have to add the “$$+\,C$$” part to it (the constant of integration), if we were to graph the solutions, we’d have many different curves that are the same but vertical translations of each other. But sometimes we know more about the curve, so we can determine a particular solution (get what the $$C$$ is), and we can do this say with one $$(x,y)$$ value on the curve, which is called the initial condition.

Then, take the integral with the “$$+\,C$$” part, and then put in the initial condition values given for $$x$$ and $$y$$ (which is $$f(x)$$), and solve for $$C$$. This will give you the particular solution.

Note that we have to go through the exercise more than once when we are given the second derivative.

Let’s do some examples:

 Indefinite Integral Problem Particular Solution Find the particular solution $$f\left( x \right)$$ if $${f}’\left( x \right)=3x+2$$ and $$f\left( 2 \right)=6$$ $$\displaystyle f\left( x \right)=\int{{\left( {3x+2} \right)}}\,dx=3\int{{x\,dx+\int{2}\,dx}}=3\cdot \frac{{{{x}^{2}}}}{2}+2x+C=\frac{3}{2}{{x}^{2}}+2x+C$$ Put in initial condition: $$\displaystyle f\left( 2 \right)=6=\frac{3}{2}{{\left( 2 \right)}^{2}}+2\left( 2 \right)+C;\,\,\,\,\,\,C=6-\left( {\frac{3}{2}{{{\left( 2 \right)}}^{2}}+2\left( 2 \right)} \right)=-4;\,\,\,\,C=-4$$ Therefore, $$\displaystyle f\left( x \right)=\frac{3}{2}{{x}^{2}}+2x-4$$ Find the particular solution $$f\left( x \right)$$ if $${f}’\left( x \right)=2{{\sec }^{2}}x+3\sin x$$ and $$f\left( \pi \right)=3$$. $$f\left( x \right)=\int{{\left( {2{{{\sec }}^{2}}x+3\sin x} \right)}}\,dx=2\int{{{{{\sec }}^{2}}x\,dx+3\int{{\sin x}}\,dx}}=2\tan x-3\cos x+C$$ Put in initial condition: $$\begin{array}{c}f\left( \pi \right)=3=2\tan \pi -3\cos \pi \,+C;\,\,\,\,C=3-\left( {2\tan \pi -3\cos \pi } \right)\,\,\\C=3-\left( {0-\left( {-3} \right)} \right);\,\,\,C=0\end{array}$$ Therefore, $$f\left( x \right)=2\tan x-3\cos x\,$$ Second derivative problem:   Find the particular solution $$f\left( x \right)$$ if $${{f}^{\prime \prime}(x)}={{x}^{3}}+3$$ and $${f}’\left( 0 \right)=-5$$ and $$f\left( 1 \right)=4$$. We have to integrate and find “C” values twice, since we’re starting with the second derivative: $$\displaystyle {f}’\left( x \right)=\int{{\left( {{{x}^{3}}+3} \right)}}\,dx=\int{{{{x}^{3}}\,dx+\int{3}\,dx}}=\frac{{{{x}^{4}}}}{4}+3x+C$$ Put in initial condition for first derivative: $$\displaystyle {f}’\left( 0 \right)=-5=\frac{{{{x}^{4}}}}{4}+3x+C;\,\,\,\,\,C=-5-\left( {\frac{{{{0}^{4}}}}{4}+3\left( 0 \right)} \right);\,\,\,\,C=-5$$ Therefore, $$\displaystyle {f}’\left( x \right)=\frac{{{{x}^{4}}}}{4}+3x-5$$ Now take integral again to get $$f\left( x \right)$$: \displaystyle \begin{align}f\left( x \right)&=\int{{\left( {\frac{{{{x}^{4}}}}{4}+3x-5} \right)}}\,dx=\frac{1}{4}\int{{{{x}^{4}}\,dx+3\int{x}\,dx}}+\int{{-5}}\,dx\\&=\frac{{{{x}^{5}}}}{{20}}+\frac{{3{{x}^{2}}}}{2}-5x\,+C\end{align} Put in initial condition: $$\displaystyle f\left( 1 \right)=4=\frac{{{{x}^{5}}}}{{20}}+\frac{{3{{x}^{2}}}}{2}-5x\,+C;\,\,\,\,C=4-\left( {\frac{{{{1}^{5}}}}{{20}}+\frac{{3{{{\left( 1 \right)}}^{2}}}}{2}-5\left( 1 \right)} \right)$$ $$\displaystyle C=4-\left( {-\frac{{69}}{{20}}} \right)=\frac{{149}}{{20}}$$ Therefore, $$\displaystyle {f}’\left( x \right)=\frac{{{{x}^{5}}}}{{20}}+\frac{{3{{x}^{2}}}}{2}-5x+\frac{{149}}{{20}}$$

And here’s a word problem. Note how in this problem, we have to use a system of equations to solve for the particular solution. Also remember when something is proportional to something else, it’s a direct variation, and one side is the product of a constant $$k$$ and the other side.

 Indefinite Integral Word Problem Solution The growth of a population of bacteria is proportional to the cube root of $$t$$, where $$P$$ is the population size and $$t$$ is the time in days.   Thus, we have:   $$\displaystyle \frac{{dP}}{{dt}}=k\sqrt[3]{t}$$   The initial size of the population is 800, and after one day the population grows to 1000.   Estimate the population $$P$$ after 1 week. Since we have the growth function (which is a rate), but we want the actual population (which is a number of bacteria), we need to integrate. Remember that the derivative is a rate of something: $$\displaystyle f\left( t \right)=\int{{\left( {k\sqrt[3]{t}} \right)}}\,dt=k\int{{\left( {{{t}^{{^{{\frac{1}{3}}}}}}} \right)}}\,dt=k\frac{{{{t}^{{^{{\frac{4}{3}}}}}}}}{{\left( {\frac{4}{3}} \right)}}+C=\frac{{3k{{t}^{{^{{\frac{4}{3}}}}}}}}{4}+C$$ We have two unknowns ($$k$$ and $$C$$), so we’ll have to come up with two equations to solve. With two data points $$\left( {0,800} \right)$$ and $$\left( {1,1000} \right)$$, this should work:   Put in initial conditions: $$\displaystyle f\left( 0 \right)=800=\frac{{3k{{{\left( 0 \right)}}^{{^{{\frac{4}{3}}}}}}}}{4}+C;\,\,\,\,800=0+C;\,\,\,C=800$$, so $$\displaystyle f\left( t \right)=\frac{{3k{{t}^{{^{{\frac{4}{3}}}}}}}}{4}+800$$ $$\displaystyle f\left( 1 \right)=1000=\frac{{3k{{{\left( 1 \right)}}^{{^{{\frac{4}{3}}}}}}}}{4}+800;\,\,\,\,1000=\frac{{3k}}{4}+800;\,\,\,\,\,200=\frac{{3k}}{4};\,\,\,\,k=200\cdot \frac{4}{3}=\frac{{800}}{3}$$ $$\require {cancel} \displaystyle f\left( t \right)=\frac{{\cancel{3}\left( {\frac{{800}}{{\cancel{3}}}} \right){{t}^{{^{{\frac{4}{3}}}}}}}}{4}+800,\,\,\text{or}\,\,f\left( t \right)=200\sqrt[3]{{{{t}^{4}}}}+800$$   To estimate the population after 1 week (7 days), we have $$f\left( 7 \right)=200\sqrt[3]{{{{7}^{4}}}}+800\approx 3478\,\,\text{bacteria}$$

# Revisiting Position, Velocity, and Acceleration

We talked a little about Rates of Change and Velocity here in the Equation of the Tangent Line, Tangent Line Approximation, and Rates of Change section (and we will here in the Definite Integration section). Now we’ll be able to work backwards (for example, from a rate to a distance) since we know how to integrate functions!

Here are some things we know about position, velocity, and acceleration:

• If a particle is moving along a horizontal line, its position (relative to say the origin) is a function, but the derivative of this function would be its (instantaneousvelocity (how fast it’s moving) at a certain point, and the derivative of its velocity would be its acceleration (how fast its velocity is changing).
• We can take the integral of velocity to get position, and the integral of acceleration to get velocity.
• The position of an object is actually a vector, since it has both a magnitude (a scalar, such as distance) and a direction. A change in position is a displacement, which is how far out of place the object is, compared to where it started. The distance it has traveled is the total amount of ground an object has covered during its motion; this is the absolute value of the displacement, and is a scalar.
• The velocity function is the derivative of the position function, and be negative, zero, or positive. If the derivative (velocity) is positive, the object is moving to the right (or up, if that’s how the coordinate system is defined); if negative, it’s moving to the left (or down); if the velocity is 0, the object is at rest. This is also called the direction of the object.
• The velocity of an object is actually a vector, whereas the speed is the absolute value of the velocity, and is a scalar. The speed of an object cannot be negative, whereas velocity can.
• Acceleration (the derivative of velocity, which is also a vector) can cause speed to increase, decrease, or stay the same. Negative acceleration means slowing down (velocity decreasing) and positive acceleration means speeding up (velocity increasing).

Let’s do some problems:

 Acceleration Integral Problem Solution A particle, starting out at rest, moves along the $$x$$-axis such that its acceleration is $$a\left( t \right)=2\cos \left( t \right)$$, $$t$$ is in seconds. At the time $$t=0$$, its position is $$p=3$$.   a) Find the velocity $$v\left( t \right)$$ of the particle at time $$t$$.   b) Find when the particle is at rest.   c) Find the position of the particle $$p\left( t \right)$$ at any time $$t$$. a) We are given the acceleration, but we need to get a formula for the velocity; we can do that now with integration: $$\displaystyle v\left( t \right)=\int{{a\left( t \right)dt;\,\,\,\,v\left(t\right)=\int{{2\cos \left( t \right)dt=2\int{{\cos \left( t \right)dt=2\sin \left( t \right)}}}}}}+C$$ To solve for $$C$$ (the constant of integration), put in the initial condition, which states that the particle is at rest when it starts out, so we have the point $$\left( {t,v\left( t \right)} \right)=\left( {0,0} \right)$$: $$v\left( 0 \right)=0=2\sin \left( 0 \right)+C;\,\,\,\,0=2\cdot 0+C;\,\,\,C=0$$ Thus, the velocity of the particle at time $$t$$ is $$\displaystyle v\left( t \right)=2\sin \left( t \right)$$.   b) A particle is at rest when the velocity is 0: $$\displaystyle v\left( t \right)=2\sin \left( t \right)=0;\,\,\,\sin \left( t \right)=0;\,\,\,t=0,\,\pi$$ (from unit circle) Thus, the particle is at rest at 0 and about 3.14 seconds.   c) Now that we have the velocity equation $$\displaystyle v\left( t \right)=2\sin \left( t \right)$$, we can get the position equation by integrating again: $$\displaystyle p\left( t \right)=\int{{v\left( t \right)dt;\,\,\,\,p\left( t \right)=\int{{2\sin \left( t \right)dt=2\int{{\sin \left( t \right)dt=-2\cos \left( t \right)}}}}}}\,+C$$ To solve for $$C$$ (the constant of integration), put in the initial condition, which states that at time $$t=0$$, its position is $$p=3$$, so we have the point $$\left( {t,p\left( t \right)} \right)=\left( {0,3} \right)$$: $$p\left( 0 \right)=3=-2\cos \left( 0 \right)+C;\,\,\,\,3=-2\cdot 1+C;\,\,\,C=5$$ Thus, the position of the particle at time $$t$$ is $$\displaystyle p\left( t \right)=-2\cos \left( t \right)+5$$. An automobile maker claims that their car can accelerate at a constant rate, and the car can accelerate from a dead stop to a velocity of 75 ft/sec in a distance of 150 feet.   What is the acceleration of the car? First, start with an acceleration formula since we’re working on integration in this section (this can be a clue!). Since the car has a constant acceleration, we have $$\displaystyle a\left( t \right)=kt$$. (We need to get $$k$$ to get the car’s acceleration). Now, integrate to get a velocity formula: $$\displaystyle v\left( t \right)=\int{{a\left( t \right)dt;\,\,\,\,v\left( t \right)=\int{{ktdt=k\int{{dt=kt}}}}}}+C$$ Since we know that the car starts from a dead stop, we have no velocity at $$\left( {t,v\left( t \right)} \right)=\left( {0,0} \right)$$, so we have: $$\displaystyle v\left( 0 \right)=0=k\cdot 0+C=0;\,\,\,\,C=0$$. Thus, the velocity of the car at time $$t$$ is $$\displaystyle v\left( t \right)=kt$$ (relating velocity, acceleration and time). We can’t do more at this point, so let’s get the position of the car $$\displaystyle p\left( t \right)=\int{{v\left( t \right)dt;\,\,\,\,p\left( t \right)=\int{{ktdt=k\int{{tdt=\frac{1}{2}k{{t}^{2}}}}}}}}+C$$ Again since the car is presumably starting at position 0, we have $$\left( {t,p\left( t \right)} \right)=\left( {0,0} \right)$$, so $$\displaystyle p\left( t \right)=\frac{1}{2}k{{t}^{2}}$$.   This is where it gets tricky; we have to relate the velocity equation with the position equation, since we know that the time the car has a speed of 75 ft/sec, the distance is 150 ft. Thus, $$\left( {p\left( t \right),v\left( t \right)} \right)=\left( {150,75} \right)$$. Solve for $$k$$ in the velocity equation: $$\displaystyle v\left( t \right)=kt,\,\,\,\text{or}\,\,k=\frac{{v\left( t \right)}}{t}$$, and throw this into the position equation and get this time: $$\require {cancel} \displaystyle p\left( t \right)=\frac{1}{2}k{{t}^{2}}=\frac{1}{2}\cdot \frac{{v\left( t \right)}}{{\cancel{t}}}{{\cancel{{{{t}^{2}}}}}^{t}}=\frac{{v\left( t \right)\cdot t}}{2}$$ Plug in $$\left( {p\left( t \right),v\left( t \right)} \right)=\left( {150,75} \right)$$ to get $$\displaystyle 150=\frac{{75t}}{2};\,\,\,t=4\text{ sec}$$, which means $$\displaystyle k=\frac{{v\left( t \right)}}{t}=\frac{{75}}{4}=18.75\,\,\text{ft/se}{{\text{c}}^{2}}$$ at $$t=4$$

Here’s one more problem:

Acceleration Integral Problem Solution
At the moment a traffic light turns green, Molly, who has been waiting at the intersection starts with a constant acceleration of 10 feet per second2.

At the same moment, Molly’s friend Quinn, who is traveling at a constant velocity of 35 feet per second passes Molly.

Molly passes Quinn sometime after this. How fast will Molly be traveling when she passes Quinn?

Where do we start? Let’s draw a chart with all the details we have for Molly and Quinn, and integrate until we get to position, which is what we need to use to compare them. This is because the question wants to know how fast Molly will be traveling when she passes (same position as) Quinn.

We will assume that the two girls have a velocity and acceleration of 0 when they are at position 0 (the intersection), so we have $$\left( {t,a\left( t \right)} \right)=\left( {t,v\left( t \right)} \right)=\left( {0,0} \right)$$; in all cases, $$C=0$$:

 Molly Quinn $$\begin{array}{c}a=10\\v\left( t \right)=\int{{a\left( t \right)}}\,dt=\int{{10}}\,dt=10t+C\\C=0;\,\,\,\,v\left( t \right)=10t\\\\p\left( t \right)=\int{{v\left( t \right)}}\,dt=\int{{10t}}\,dt=5{{t}^{2}}+C\\C=0;\,\,\,\,p\left( t \right)=5{{t}^{2}}\end{array}$$ $$\begin{array}{c}v\left( t \right)=35\\\\p\left( t \right)=\int{{v\left( t \right)}}\,dt=\int{{35}}\,dt=35t+C\\\,C=0;\,\,\,\,\,\,p\left( t \right)=35t\end{array}$$

Put the two position formulas together and solve for $$t$$, which is the time that they are at the same position:

$$5{{t}^{2}}=35t;\,\,\,\,5t\left( {t-7} \right)=0;\,\,\,\,t=0,\,\,7\,\,\,\text{sec}$$

The times they will be at the same position is 0 seconds (makes sense; they haven’t left yet) and 7 seconds, when Molly passes Quinn. But the question is asking how fast will Molly be traveling (velocity) when she passes Quinn. She will be traveling at $$v=10t=10\left( 7 \right)=70\,$$ feet per second.

Again, make sure you check exactly what the question is asking for, which is velocity, not time!

Here are a few more, taken from Calculus, 10th Edition, by Ron Larson and Bruce Edwards:

 Acceleration Integral Problem Solution The speed of a car traveling in a straight line is reduced from 45 to 30 miles per hour in a distance of 264 feet.     Find the distance in which the car can be brought to rest from 30 miles per hour, assuming the same constant deceleration.     First use dimensional analysis to convert the velocities to feet, since the problem asks for the distance, which is in feet: \require {cancel} \begin{align}45\frac{{\cancel{{\text{miles}}}}}{{\cancel{{\text{hour}}}}}\cdot \frac{{5280\,\text{feet}}}{{\text{1 }\cancel{{\text{mile}}}}}\cdot \frac{{1\,\cancel{{\text{hour}}}}}{{3600\,\,\text{seconds}}}&=66\frac{{\text{feet}}}{{\text{second}}}\\30\frac{{\cancel{{\text{miles}}}}}{{\cancel{{\text{hour}}}}}\cdot \frac{{5280\,\text{feet}}}{{\text{1 }\cancel{{\text{mile}}}}}\cdot \frac{{1\,\cancel{{\text{hour}}}}}{{3600\,\,\text{seconds}}}&=44\frac{{\text{feet}}}{{\text{second}}}\end{align} Start with acceleration and integrate to get velocity. At time 0, the velocity is 66 feet per second: \begin{align}a(t)&=a\\v\left(t \right)&=\int{{a\left( t \right)}}=at+C\\66&=a\left( 0 \right)+C;\,\,\,C=66\\v\left( t \right)&=at+66\end{align}   Integrate to get position, assuming that at $$t=0$$, we have $$p=0$$, so $$C=0$$: $$\displaystyle p\left( t \right)=\int{{v\left( t \right)=}}\frac{1}{2}a{{t}^{2}}+66t+C;\,\,\,p(t)=\frac{1}{2}a{{t}^{2}}+66t$$   Use the fact that the time when the velocity is 44 feet per second is the same as when the position is 264 feet (we can connect these two equations with time): \begin{align}44&=at+66\\264&=\frac{1}{2}a{{t}^{2}}+66t\end{align}   Solve the system of equations (I did on a graphing calculator) to get $$t=4.8\,\sec ,\,\,a=-4.5833\,\text{feet}\,\text{per}\,\text{secon}{{\text{d}}^{2}}$$. We want the distance in which the car can be brought to rest (velocity is 0 feet per second); start filling in the equations: $$\begin{array}{l}a(t)=-4.5833\\v\left( t \right)=-4.5833t+66\\s\left( t \right)=-2.2917{{t}^{2}}+66t\end{array}$$   When velocity is 0, $$\displaystyle 0=-4.5833t+66;\,\,\,t=14.4\,\text{sec}$$. The position at that time is $$s\left( {14.4} \right)=-2.2917{{\left( {14.4} \right)}^{2}}+66\left( {14.4} \right)=475.2\,\text{feet}$$. But the car has already gone 264 feet, so the rest of the distance is $$475.2-264=211.2\,\text{feet}$$. Tricky! An airplane taking off from a runway travels 3600 feet before lifting off.     The airplane starts from rest, moves with constant acceleration, and makes the run in 30 seconds.     With what speed does it lift off? Start with acceleration and integrate to get velocity and position. At time 0, the velocity is 0 feet per second: \begin{align}a(t)&=a\\v\left( t \right)&=\int{{a\left( t \right)}}=at+C\\0&=a\left( 0 \right)+C;\,\,\,C=0\\v\left( t \right)&=at\\p\left( t \right)&=\int{{v\left( t \right)=}}\frac{1}{2}a{{t}^{2}}+C;\,\,C=0\\p\left( t \right)&=\frac{1}{2}a{{t}^{2}}\end{align}   Since the airplane makes the 3600 feet run in 30 seconds, we can use the position formula: \begin{align}p\left( t \right)&=\frac{1}{2}a{{t}^{2}}\\3600&=\frac{1}{2}a{{\left( {30} \right)}^{2}}\\a&=8\,\text{feet}\,\text{per}\,\text{secon}{{\text{d}}^{2}}\end{align} We want the speed at that time: \displaystyle \begin{align}v\left( t \right)&=at\\v\left( 8 \right)=8\left( {30} \right)&=240\,\text{feet per second}\end{align} Not too bad!

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On to U-Substitution Integration – you’re ready!