Integration - AP Calculus BC

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Question

$\int \frac{2t-1}{t^2+4}dt$

Answer

At first, we notice that the power in the numerator is smaller than the one in the denominator, but we can't quite use u-substitution yet, as the derivative of the denominator is 2t - not 2t -1. To do this we must separate the integrals:

$=\int \frac{2t}{t^2+4}dt - \int \frac{1}{t^2+4}dt$

Now, we can use u-substitution for the first integral by setting

$dt = \frac{du}{2t}$

Which results in:

$\int \frac{2t}{t^2-4}dt = \int \frac{2t}{u}(\frac{du}{2t})$

The 2t's cross out and now we get:

$\int \frac{2t}{t^2-4}dt = \int \frac{1}{u}du$

Now integrate as normal:

$\int \frac{2t}{t^2-4}dt = ln\left | u\right |+C$

Substitution back for , we now get:

$\int \frac{2t}{t^2-4}dt = ln\left | t^2+4\right |+C$

For the second integral, $\int \frac{1}{t^2+4}dt$ , we have to use a special formula:

$\int \frac{1}{x^2+a^2}dx = \frac{1}{a}tan^{-1} (\frac{x}{a})+C$ , or in this case translates to:

$\int \frac{1}{t^2+4}dt = \frac{1}{2}tan^{-1}(\frac{t}{2})+C$

Combining the two results get us:

$\int \frac{2t-1}{t^2+4}dt = ln\left | t^2+4\right|-\frac{1}{2}\tan^{-1}\frac{t}{2}+C$

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Question

Function gives the velocity of a particle as a function of time.

$\small \small v(t)=t^5+3t^4-\frac{7t^2}{3}$

Find the equation that models the particle's postion as a function of time.

Answer

Recall that velocity is the first derivative of position, and acceleration is the second derivative of position. We begin with velocity, so we need to integrate to find position and derive to find acceleration.

We are starting with the following

$\small \small v(t)=t^5+3t^4-\frac{7t^2}{3}$

We need to perform the following:

$\small \small \small \small\int v(t)dt=\int t^5+3t^4-\frac{7t^2}{3}dt$

Recall that to integrate, we add one to each exponent and divide by the that number, so we get the following. Don't forget your +c as well.

$\small \small \int t^5+3t^4-\frac{7t^2}{3}dt=\frac{t^6}{6}+\frac{3t^5}{5}-\frac{7t^3}{9}+c$

Which makes our position function, h(t), the following:

$\small h(t)=\frac{t^6}{6}+\frac{3t^5}{5}-\frac{7t^3}{9}+c$

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Question

Consider the velocity function modeled in meters per second by v(t).

Find the position of a particle whose velocity is modeled by after seconds.

Answer

Recall that velocity is the first derivative of position, so to find the position function we need to integrate .

$V(t)=\int 4t^3-5t^2+6dt$

Becomes,

$V(t)=t^4-\frac{5t^3}{3}+6t+c$

Then, we need to find

$V(10)=10^4-\frac{5\cdot10^3}{3}+6\cdot10+c=8393.3\bar{3}+c$

So our final answer is:

$8393.3\bar{3}+c \text{ }\frac{m}{s}$

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Question

Find the position function given the velocity function:

Answer

To find the position from the velocity function, integrate

by increasing the exponent of each t term and then dividing that term by the new exponent value.

$s(t)=\int v(t)dt= \frac{3}{6}t^6+2t^2=\frac{1}{2}t^6+2t^2$

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Question

If the acceleration function of an object is , what is the position of the object at ? Assume the initial velocity and position is zero.

Answer

To find the position function from the acceleration function, integrate twice.

$v(t)=\int a(t)dt=t^2+11t$

When integrating, remember to increase the exponent of the variable by one and then divide the term by the new exponent. Do this for each term.

$s(t)=\int v(t)dt=\int\int a(t)dt = \frac{1}{3}t^3+\frac{11}{2}t^2$

Solve for .

$\frac{1}{3}(1)^3+\frac{11}{2}(1)^2=\frac{1}{3}+\frac{11}{2} = \frac{2}{6}+\frac{33}{6}= \frac{35}{6}$

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Question

Consider the velocity function given by :

Find the position of a particle after seconds if its velocity can be modeled by and the graph of its position function passes through the point .

Answer

Recall that velocity is the first derivative of position and acceleration is the second derivative of position. Therfore, we need to integrate v(t) to find p(t)

$p(t)=V(t)=\int 5t^4-343t^6+51t^2dt$

So we get:

What we ultimately need is p(5), but first we need to find c: Use the point (2,2)

$2=2^5-49\cdot 2^7+17\cdot 2^3+c$

So our position function is:

$p(5)=5^5-49\cdot 5^7+17\cdot 5^3+6106=-3816769$

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Question

The velocity equation of an object is given by the equation . What is the position of the object at time if the initial position of the object is ?

Answer

The position of the object can be found by integrating the velocity equation and solving for . To integrate the velocity equation we first rewrite the equation.

To integrate this equation we must use the power rule where,

$f'(x) = x^n \rightarrow f(x) = \frac{x^{n+1}}{n+1}$ .

Applying this to the velocity equation gives us,

$x(t) = \int (12t^3)dt = \frac{12t^{3+1}}{3+1} = \frac{12t^4}{4} = 3t^4 +C$ .

We must solve for the value of by using the initial position of the object.

Therefore, and .

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Question

The acceleration of an object is given by the equation . What is the equation for the position of the object, if the object has an initial velocity of and an initial position of ?

Answer

To find the position of the object we must use the power rule to integrate the acceleration equation twice. The power rule is such that

$f'(x) = x^n \rightarrow f(x) = \frac{x^{n+1}}{n+1}$

Therefore integrating the acceleration equation gives us

$v(t) = \int (4t+1)dt = \frac{4t^{1+1}}{1+1} +\frac{1t^{0+1}}{0+1} +C = 2t^2 + t +C$

We can solve for the value of by using the initial velocity of the object.

Therefore and

To find the position of the object we integrate the velocity equation.

$x(t) = \int (2t^2+t+10)dt = \frac{2t^{2+1}}{2+1} + \frac{t^{1+1}}{1+1} + \frac{10t^{0+1}}{0+1} +C = \frac{2t^3}{3}+\frac{t^2}{2}+10t +C$

We can solve for this new value of by using the object's initial position

$x(0) = \frac{2(0)^3}{3} + \frac{(0)^2}{2} + 10(0) +C = 0 + C = 0$

Therefore and $x(t) = \frac{2t^3}{3} + \frac{t^2}{2} + 10t$

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Question

The velocity of an object is given by the equation . What is the position of the object at time if the object has a position of and time ?

Answer

To find the position of the object we must first find the position equation of the object. The position equation can be found by integrating the velocity equation. This can be done using the power rule where if

$f'(x) = x^n \rightarrow f(x) = \frac{x^{n+1}}{n+1}$

Using this rule we find that

$x(t) = \int (6t-3)dt = \frac{6t^{1+1}}{1+1} - \frac{3t^{0+1}}{0+1} +C = 3t^2 - 3t +C$

Using the position of the object at time we can solve for

Therefore and

We can now find the position at time .

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Question

The velocity of an object is . What is the position of the object if its initial position is ?

Answer

The position is the integral of the velocity. By integrating with the power rule we can find the object's position.

The power rule is where

$f'(x) = x^n \rightarrow f(x) = \frac{x^{n+1}}{n+1}$ .

Therefore the position of the object is

$x(t) = \int (3t^2)dt = \frac{3t^{(2+1)}}{2+1} +C = t^3 +C$ .

We can solve for the constant using the object's initial position.

Therefore and .

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Question

If the velocity function of a car is , what is the position when ?

Answer

To find the position from the velocity function, take the integral of the velocity function.

$\int v(t):dt=\int 3cos(t):dt$

Substitute .

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Question

If the velocity of a particle is $\small v(t)=t^3-3t^2$ , and its position at $\small \small t=2$ is $\small \small \small s(2)=4$ , what is its position at $\small \small t=4$ ?

Answer

This problem can be done using the fundamental theorem of calculus. We hvae

$\small \small \int_2^4 v(t)dt=\int_2^4 s'(t)dt=s(4)-s(2)$

where $\small s(t)$ is the position at time $\small t$ . So we have

$\small \small \small s(4)=s(2)+\int_2^4 v(t)dt$

so we need to solve for $\small \small \small \small \int_2^4 v(t)dt$ :

$\small \small \small \small \small \small \int_2^4 v(t)dt=\int_2^4 (t^3-3t^2)dt=\frac{t^4}{4}-t^3|_2^4$

$\small =\left(\frac{4^4}{4}-4^3\right)-\left(\frac{2^4}{4}-2^3\right)=(4^3-4^3)-\left(\frac{16}{4}-8\right)=-(4-8)=4$

So then the position at $\small t=4$ is

$\small \small s(4)=s(2)+\int_2^4 v(t)dt=4+4=8$

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Question

A particle's velocity in two dimensions is described by the functions:

If the particle has an initial position of , what will its position be at time $t=2\pi$ ?

Answer

Position can be found by integrating velocity with respect to time:

$p(t)=\int v(t)dt$

For velocities:

The position functions are:

$x(t)=\frac{t^3}{3}-cos(t)+C_x$

$y(t)=\frac{sin(t^2)}{2}+3t+C_y$

These constants of integration can be found by using the given initial conditions:

$x(0)=\frac{0^3}{3}-cos(0)+C_x=0$

$y(0)=\frac{sin(0^2)}{2}+3(0)+C_y=0$

Which gives the definite integrals:

$x(t)=\frac{t^3}{3}-cos(t)+1$

$y(t)=\frac{sin(t^2)}{2}+3t$

$x(2\pi)=\frac{(2\pi)^3}{3}-cos(2\pi)+1$

$x(2\pi)=\frac{8\pi^3}{3}$

$y(2\pi)=\frac{sin((2\pi)^2)}{2}+3(2\pi)$

$y(2\pi)=6\pi$

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Question

Calculate antiderivative of $f(x)=\frac{3x^2+2x+1}{x^3+x^2+x+1}$

Answer

We can calculate the antiderivative $\small f(x)=\frac{3x^2+2x+1}{x^3+x^2+x+1}$

$\small \small \int \frac{3x^2+2x+1}{x^3+x^2+x+1}dx$

by using $\small u$ -substitution. We set $\small u=x^3+x^2+x+1$ , so we get $\small du=3x^2+2x+1$ , so the integral becomes

$\small \small \small \int \frac{3x^2+2x+1}{x^3+x^2+x+1}dx=\int \frac{du}{u}=\ln u$

So we just plug $\small u=x^3+x^2+x+1$ to get

$\small \small \small \small \small \int \frac{3x^2+2x+1}{x^3+x^2+x+1}dx=\int \frac{du}{u}=\ln u=\ln (x^3+x^2+x+1)$

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Question

Calculate antiderivative of $\small f(x)=ex$ .

Answer

We can calculate the antiderivative $\small \small f(x)=ex$

$\small \small \small \int ex dx$

by using the power rule for antiderivatives:

$\small \int x^a=\frac{x^{a+1}}{a+1}+C$

In this case $\small a=1$ , so we have

$\small \small \small \small \small \int ex dx=e\frac{x^2}{2}+C=\frac{ex^2}{2}+C$ .

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Question

The physical interpretation of the integral of a function , denoted by , is what?

Answer

By definition, the integral of a fucntion is the summation of an infinite number of small areas, thus giving the total area of the function with respect to the axis of integration.

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Question

Calculate the integral of the function given below.

$f (x) = x^4 - \cos (2x)$

Answer

We have a seperable integral, meaning each term can be integrated independently, written as

$\int f (x)dx= \int \left(x^4 - \cos (2x)\right )dx = \int x^4 dx - \int \cos(2x)dx$ .

The first term is a power-rule integral, while the second is a trigonometric intergral. One should recall

$\int x^n dx = \frac{1}{n+1}x^{n+1}+C$

and $\int \cos (ax)dx = \frac{1}{a} \sin(ax)+C$

Putting these facts together leads us to the final answer of

$\int f (x)dx= \frac{x^5}{5}-\frac{1}{2} \sin (2x)+C$ .

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Question

Evaluate $\int_0^1 2t\sqrt{1+t^2}dt$ .

Answer

The correct answer is $\frac{2}{3}(2^{3/2}-1)$ .

We can evaluate this integral using -substitution.

Let , then , hence we have

$\int_0^1 2t\sqrt{1+t^2}dt$ (Don't forget to change the bounds of integration)

$=\int_1^2 \sqrt{u} du$

$= \int_1^2 u^{1/2}du$

$=[\frac{2u^{3/2}}{3}]^2_1$

$=\frac{2\times{2}^{3/2}}{3} -\frac{2}{3}$

$=\frac{2}{3}(2^{3/2}-1)$ .

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Question

Evaluate $\int_0^t e^ax dx$ , where is any constant.

Answer

Since is a constant, so is , therefore we can factor it out of the integral.

$\int_0^te^ax dx$

$=e^a\int_0^txdx$

$=e^a[\frac{x^2}{2}]^t_0$

$=e^a(\frac{t^2}{2}-0)$

$=\frac{t^2e^a}{2}$

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Question

Calculate $\int \cos (x) e^{sin(x)}dx$ .

Answer

This integral can be found using u-substitution. Consider

$u = \sin(x), du = \cos(x)$ . This means we can rewrite our integral as

$\int e^{u}du=e^u+C$ , by definition of the integral of an exponential.

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