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AP Calculus AB › Equations

Questions 1 - 10

Write the correct expression to find the area of from .

$\int_{-2}^{0}-xdx+\int_{0}^{2}xdx$

$\int_{-2}^{2}xdx$

$2\int_{-2}^{2}xdx$

$\int_{-2}^{2}-xdx+\int_{-2}^{2}xdx$

$\textup{The correct answer is not given.}$

Explanation

This is a tricky question.

If we chose $\int_{-2}^{2}xdx$ , the region in the third quadrant of is negative area, since is no longer the top curve. Evaluating this integral will cancel out the negative area in the third quadrant with the positive area on the first quadrant. This integral will give zero area, which is incorrect.

The correct method is to split this interval into 2 separate integrations: One from and the other from .

From interval , the top curve is minus the bottom curve, , is .

Set up the integral.

$\int_{-2}^{0}-xdx+\int_{0}^{2}xdx$

Solve the following integral, where a and b are constants:

$\int abe^{b^2x}dx$

$\frac{ae^{b^2x}}{b}+C$

$ae^{b^2x}+C$

$\frac{2ae^{b^2x}}{b}+C$

$\frac{1}{2}ae^{b^2x}+C$

Explanation

Keeping in mind that a and b are only constants, the integral is equal to

$\int abe^{b^2x}dx=\frac{abe^{b^2x}}{b^2}+C=\frac{ae^{b^2x}}{b}+C$

and was found using the following rule:

$\int e^{ax}dx=\frac{e^{ax}}{a}+C$

Take the indefinite integral of

$\int x(x^6+7)^2dx$

$\frac{1}{14}x^{14}+\frac{7}{4}x^8+\frac{49}{2}x^2+C$

$x^2(x^6+7)^3+C$

$\frac{1}{13}x^{13}+2x^7+49x+C$

$(x^2+7)(x^6+7)^2$

Indefinite integral does not exist

Explanation

First, foil the integral so it is easier to manage.

$\int (x^{13}+14x^7+49x) dx$

then perform the indefinite integral the normal way you would do

$\frac{1}{14}x^{14}+\frac{7}{4}x^8+\frac{49}{2}x^2+C$

Evaluate the definite integral within the interval $4\leq x \leq 9$

$\int \left(\frac{1}{\sqrt{x}}+3x^2-4\right)dx$

$\frac{1169}{6}$

$\frac{-1169}{6}$

Explanation

In order to solve this problem we must remember that:

$\int_{a}^{b}f(x)=F(b)-F(a)$

In this case we have:

$\int_{4}^{9}\left(\frac{1}{\sqrt{x}}+3x^2-4\right)dx$

Our first step is to integrate:

$\int\left(\frac{1}{\sqrt{x}}+3x^2-4\right)dx=2\sqrt{x}+x^{3}-4x$

We then arrive to our solution by plugging in our values

$\left [ 2\sqrt{9}+(9)^{3}-4(9) \right ]-\left [ 2\sqrt{4}+(4)^{3}-4(4) \right ]$

What is the equation of the tangent line at x = 15 for f(x) = x4 + 5x2 + 44x – 3?

y = 55x + 13382

y = 2848x + 34

y = 153103x – 13694

None of the other answers

y = 13694x – 153003

Explanation

First we must solve for the general derivative of f(x) = f(x) = x4 + 5x2 + 44x – 3.

f'(x) = 4x3 + 10x + 44

Now, the slope of the tangent line for f(15) is equal to f'(15):

f'(15) = 4(15)3 + 10 *15 + 44 = 13694.

To find the tangent line, we need at least one point on the line. To find this, we can use f(15) to get the y value of the point of tangency, which will suffice for our use:

f(15) = 154 + 5(15)2 + 44 * 15 – 3 = 50625 + 1125 + 660 – 3 = 52407

Now, using the point-slope form of the line, we get:

y - 52407 = 13694 * (x – 15)

Simplify:

y – 52407 = 13694x – 205410

y = 13694x – 153003

Given the one-to-one equation f(x)=3x+1, the inverse function f-1(y)=

3x-1

(x-1)/3

(y+1)/3

(y-1)/3

undefined

Explanation

Before we solve the problem by computation, let's look at the answer choices and see if we can eliminate any answer choices. We know that an inverse function must exist because f(x) is one-to-one, so we can eliminate the answer choice "undefined." Next, we know that the inverse function has to be in terms of y, so we can eliminate the two answer choices with an "x."

Now we can look at the two remaining answer choices. Let y=f(x) and solve for x to find the inverse.

So f(x)=y=3x+1. Solve for x.

x=(y-1)/3

Therefore our answer is (y-1)/3.

$\int_{0}^{2}2x^2-5x+1 dx$

$-\frac{8}{3}$

$\frac{8}{3}$

$-\frac{5}{3}$

$-\frac{11}{3}$

Explanation

When integrating, remember to add one to the exponent and then put that result on the denominator: $\frac{2x^3}{3}-\frac{5x^2}{2}+x$ . Now evaluate at 2, and then 0. Then subtract the two results. $(\frac{16}{3}-10+2)-0=-\frac{8}{3}$ .

Evaluate the following integral:

$\int x^3\cos(x^4)dx$

$\frac{1}{4}\sin(x^4)+C$

$\sin(x^4)+C$

$-\frac{1}{4}\sin(x^4)+C$

$\frac{1}{4}\sin(x^4)$

Explanation

To integrate, we must make the following subsitution:

We used the following rule to derivate:

$\frac{\mathrm{d} }{\mathrm{d} x}x^n=nx^{n-1}$

Now, after rearranging, we get the following integral:

$\frac{1}{4}\int \cos(u) du$

and after integrating we get

$\frac{1}{4}\sin(u)+C$

We used the following rule to integrate:

$\int \cos(x)dx=\sin(x)+C$

To finish, we plug our x term back in place of u:

$\frac{1}{4}\sin(x^4)+C$

Evaluate the following indefinite integral:

$\int(8x^{3}+3x^{2}+2x)dx$

$2x^{4}+x^{3}+x^{2}+C$

$2x^{4}+3x^{3}+2x+C$

$x^{4}+x^{3}+x^{2}+C$

$2x^{5}+x^{4}+3x^{2}+C$

Explanation

To evaluate the integral the integral, use the inverse power rule:

$\int x^{n}dx=\frac{x^{n+1}}{n+1} n\neq1$

Applying that rule to this problem gives us the following for the first term:

$\int 8x^{3}=\frac{8x^{3+1}}{3+1}=\frac{8x^{4}}{4}=2x^{4}$

The following for the second term:

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

And the following for the third term:

$\int 2x=\frac{2x^{1+1}}{1+1}=\frac{2x^{2}}{2}=x^{2}$

We can combine these terms and add our "C" to get the final answer:

$2x^{4}+x^{3}+x^{2}+C$

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

$ln\left | x^3+2x \right |+C$

$ln\left | x^3+2x \right |$

$ln\left | u \right |+C$

$x^3+2x \right |+C$

$\frac{ln(x^2)}{4}+C$

Explanation

To integrate this expression, you must use "u" substitution. The expression you assign to "u" is usually the expression with the higher exponent; in this case, . Since , the next step is to find du so that you can fully substitute everything.

The derivative of our u expression is

, or .

Now we can fully plug into our integral expression so we can integrate. You can rewrite the expression so it looks like this:

$\int \frac{1}{u}du$ .

Everything in the u terms fully replaced everything in the original expression. Now, we can integrate. When we integrate $\frac{1}{u}$ , it becomes $ln\left | u \right |$ .

Since it's an indefinite expression, we must add +C at the end. The last step is to sub back in the expression that "u" represents.

Therefore, our final answer is

$ln\left | x^3+2x \right |+C$ .

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