AP Calculus BC - Integration

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

$\frac{3\pi}{2}$

$-\frac{9\pi}{2}$

Explanation

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

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

Substitute .

Calculate $\int_0^{\infty} 2xe^{-x} dx$

$+ \infty$

$\frac{1}{2}$

Explanation

This can be a challenging integral using standard methods. However, it is easy if we use integration by-parts, given as

$\int udv = uv - \int v du$

Choose

$u = x, dv = e^{-x} \Rightarrow du = dx, v = -e^{-x}$ .

From the definition,

$2\int _0^{\infty}xe^{-x}dx= 2\left(-x*e^{-x} \left | _0 ^{\infty}+\int _0^{\infty}e^{-x}dx\right)=2\left(-e^{-x} \left| _0^{\infty}\right) = 2$

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

$\ln(x^3+x^2+x+1)+C$

$\ln(x^3+x^2+x)+C$

$\ln(x^3+x^2+x+20)+C$

$\ln(x^3+x^2+1)+C$

Explanation

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

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

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 ?

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

$x(t) = \frac{2t^3}{3} + \frac{t^2}{2} + 5t$

Explanation

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

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 ?

Explanation

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 .

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 .

Explanation

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:

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

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

$e^{\sin(x)}+C$

$e^{\sin (x)}$

$e^{\cos(x)}+C$

$e^{-\sin(x)}+C$

Explanation

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.

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

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

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

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

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

$ln\left ( t^2+4\right)-\frac{1}{2}\tan^{-1}\frac{t}{2}+C$

Explanation

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:

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

Evaluate the following integral:

$\int\sin^3 x\cot^2 x dx$

$-\frac{1}{3}\cos^3 x+C$

$\frac{1}{3}\cos^3 x+C$

$-\frac{1}{3}\cos^3 x$

$-\cos x+C$

Explanation

The integrand can be simplified to make the problem much easier to solve. You must use the fact that $\cot^2 x=\frac{\cos^2 x}{\sin^2 x}$ in order to rewrite the integrand. After rewriting, the problem is now:

$\int\sin x\cos^2 x dx$

Here, a simple u-substitution can be used:

$u=\cos x$ , $du=-\sin x dx$ , $dx=-\frac{du}{\sin x}$

Once again, the integrand can be rewritten with values of u:

$\int\sin x\cdot u^2\cdot -\frac{du}{\sin x}=-\int u^2du=-\frac{1}{3}u^3 +C$

The integral was taken by using the following rule:

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

Once u has been replaced with x, the final answer is:

$-\frac{1}{3}\cos^3 x+C$

Calculate $\int \cos \theta \left( 2 \sin \theta+5\right )^3 d \theta$

$\frac{1}{8}\left(2\sin \theta +5 \right )^4+C$

$\frac{\cos \theta}{8}\left(2\sin \theta +5 \right )^4+C$

$\frac{1}{4}\left(2\sin \theta +5 \right )^4+C$

$\frac{1}{8}\left(2\sin \theta +5 \right )^4- \cos \theta + C$

Explanation

This integral is most easily found by implementing u-substitution. Choose

$u = 2 \sin \theta + 5 \Rightarrow \frac{du}{2} = \cos \theta d \theta$ , which means we can rewrite the integral in a more familiar form

$\int \cos \theta \left( 2 \sin \theta + 5\right )^3 d \theta = \frac{1}{2}\int u^3 du= \frac{1}{2}*\frac{1}{4}u^4+C= \frac{1}{8}\left(2 \sin \theta + 5 \right )^4 + C$

Integration

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