Integrals

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AP Calculus BC › Integrals

Questions 1 - 10

Evaluate.

$\int x^3 - \cos (2x) \ dx$

$F(x) = \frac{x^4}{4} - \frac{\sin (2x)}{2}+C$

$F(x) = x^4 - \sin (2x)$

$F(x) = \frac{x^3}{4} - \frac{\sin (x)}{2}+C$

$F(x) = x^4 + \frac{\sin (2x)}{2}$

Answer not listed

Explanation

$\int f(x) \ dx$

In order to evaluate this integral, first find the antiderivative of

If then

If then $F(x) = \frac{x^{a+1}}{a+1} + C$

If $f(x) = \frac{1}{x}$ then

If then

If $f(x) = \cos(x)$ then $F(x) = \sin(x) + C$

If $f(x) = \sin(x)$ then $F(x) = - \cos(x) + C$

If $f(x) = \sec^2 (x)$ then $F(x) = \tan(x) + C$

In this case, $f(x) = x^3 - \cos (2x)$ .

The antiderivative is $F(x) = \frac{x^4}{4} - \frac{\sin (2x)}{2}+C$ .

Give the indefinite integral:

$\int \frac{\ln 3x ; \mathrm{d}x}{x}$

$\frac{\left ( \ln 3x \right )^2}{2} + C$

$\frac{3\left ( \ln 3 x \right )^2}{2} + C$

$\frac{x}{3}+C$

$\frac{3}{x}+C$

$\frac{ \ln 3x }{2} + C$

Explanation

Let $u = \ln 3x$ . Then $\frac{\mathrm{d} u}{\mathrm{d} x} = \frac{\mathrm{d} }{\mathrm{d} x} \ln 3x = \frac{3}{3x} = \frac{1}{x}$

and

$\mathrm{d}u = \frac{\mathrm{d}x }{x }$ .

The integral can be rewritten as

$\int u ; \mathrm{d}u$

$=\frac{ u^2}{2} + C$

$=\frac{\left ( \ln 3x \right )^2}{2} + C$

Please solve the following integral:

$\int\frac{dx}{x(ln(x))^{2}}$

$-\frac{1}{ln(x)} + C$

$\frac{1}{ln(x)} + C$

$\frac{-2}{x(ln(x))^{3}} + C$

$\frac{1}{x^{2}ln(x)} + C$

$\frac{-1}{x^{2}ln(x)} + C$

Explanation

We know that the derivative of is $\frac{1}x{}$ .

Doing a substitution and setting

$u = \ln x$ and $du = \frac{dx}{x}$ allows us to rewrite the integral as

$\int \frac{du}{u^{2}}$ which can be rewritten as $\int u^{-2}du$ .

Integrating this gets you $\frac{-1}{u}$ plus a constant (which is stated in the original question that you can assume that we already have one). Substituting back in gives us the final answer, which is

$\frac{-1}{ln(x)} + C$ .

$\int x^2e^{3x^3}dx$

$\frac{1}{9}e^{3x^3}+C$

$9e^{3x^3}+C$

$-9e^{3x^3}+C$

$3x^3e^{3x^3}+C$

$-\frac{1}{9}e^{3x^3}+C$

Explanation

Evaluating this integral requires use of a substitution. Generally with an integral involving exponential functions, we should try setting u equal to the function that e is being raised to, like so:

$\int x^2e^{3x^3}dx\rightarrow u=3x^3, du=9x^2dx\rightarrow\frac{1}{9}du=x^2dx$

Making these substitutions with the given integral we arrive at:

$\frac{1}{9}\int e^u du=\frac{1}{9}e^u+C\rightarrow\frac{1}{9}e^{3x^3}+C$

This is one of the answer choices.

$\int_{2}^{5}x^2+4x-1dx$

Explanation

First, integrate this expression. Remember to add one to the exponent and then put that result on the denominator:

$\frac{x^3}{3}+\frac{4x^2}{2}-x=\frac{x^3}{3}+2x^2-x$ .

Then, evaluate first at 5 and then at 2. Subtract those two results:

$(\frac{125}{3}+50-5)-(\frac{8}{3}+8-2)=\frac{117}{3}+39=78$ .

Thus, your answer is 78.

$\int_{1}^{3}x^3-2x^2+4 dx$

$\frac{32}{3}$

$\frac{58}{3}$

$\frac{5}{3}$

$\frac{7}{3}$

$\frac{59}{3}$

Explanation

First, integrate this expression. Remember to add one to the exponent and then put that result on the denominator:

$\frac{x^4}{4}-2(\frac{x^3}{3})+4x$

Next, evaluate at 3 and then 1. Subtract the two results:

$(\frac{81}{4}-2(9)+12)-(\frac{1}{4}-\frac{2}{3}+4)$

Simplify to get your answer:

$(\frac{81}{4}+\frac{5}{12}-6-4)=\frac{32}{3}$ .

Evaluate.

$\int 3 \ dx$

$F(x) = \frac{1}{3}x+ C$

Answer not listed

Explanation

$\int f(x) dx$

In order to evaluate this integral, first find the antiderivative of

If then

If then $F(x) = \frac{x^{a+1}}{a+1} + C$

If $f(x) = \frac{1}{x}$ then

If then

If $f(x) = \cos(x)$ then $F(x) = \sin(x) + C$

If $f(x) = \sin(x)$ then $F(x) = - \cos(x) + C$

If $f(x) = \sec^2 (x)$ then $F(x) = \tan(x) + C$

In this case, .

The antiderivative is .

Use a partial fraction decomposition to evaluate the indefinite integral,

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

$\small \ln(x-1)+2\ln(x+2)+C$

$\small \small 3\ln\left(\frac{x+2}{x-1} \right ) + C$

$\small \small \small \frac{2}{x-1}+\frac{1}{x+2} + C$

$\small \small \ln(x-1)-\ln(x+2)+C$

Explanation

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

First factor the denominator and then write the results as a sum of two fractions as follows,

$\small \small \small \frac{3x}{x^2+x-2}=\frac{3x}{(x-1)(x+2)}=\frac{A}{x-1}+\frac{B}{x+2}$ (1)

Here we have applied the reverse of the cross multiplication process. Now we carefully deduce the numerators $\small A$ and $\small B$ . Cross multiply as follows:

$\small \frac{A}{x-1}+\frac{B}{x+2}=\frac{A(x+2)+B(x-1)}{(x-1)(x+2)}$ (2)

Looking at Equation (1) and Equation (2) it's clear that we must find values of $\small A$ and $\small B$ such that:

$\small A(x+2)+B(x-1)=3x$ (3)

Note that Equation (3) is an identity, meaning it must be true for all values of $\small x.$ Therefore, we can freely choose a value of $\small x$ to eliminate either $\small A$ or $\small B$ . Choosing eliminates .

$\small \small x =1 ; ; ; ; \Rightarrow; ; ; ;A(3)=3 ; ; ; ; \Rightarrow ; ; ; ; A=1$

$\small \small x=-2 ; ; ; ; \Rightarrow ; ; ; ;B(-3)=-6; ; ; ; \Rightarrow ; ; ; ; B = 2$ Alternatively, $\small A$ and $\small B$ could have been deduced by expanding the left side of Equation (3) and collecting like terms as follows,

$\small Ax+2A+Bx-B =3x$

$\small (A+B)x+2A-B =3x$ (4)

Upon inspection of Equation (4) we see that the two following conditions for $\small A$ and $\small B$ are required,

$\small A+B = 3$

$\small 2A-B =0$

Solve for $\small A$ and $\small B$ , this can be done quickly by adding the two equations above,

$\small 3A=3 : : : \rightarrow : : :A=1$

Using this value, $\small \small A=1$ gives $\small B = 2$ .

Summarizing our partial fraction decomposition:

$\small \small \small \frac{3x}{x^2+x-2}=\frac{A}{x-1}+\frac{B}{x+2} =\frac{1}{x-1}+\frac{2}{x+2}$

Now we can use the partial fraction decomposition to compute the integral.

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

We can apply the sum rule on the right side and recognize that each term is the derivative of a natural logarithm,

$\small \small \small \int \frac{1}{x-1}dx +\int\frac{2}{x+2}dx = \ln(x-1)+2\ln(x+2)+C$

We could use the properties of logarithms to write the result as a single logarithm, but we will leave it in the form of a sum here, remembering the constant of integration .

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

Now we can check to make sure the result is correct by differentiating the result:

$\frac{d}{dx}\left[\ln(x-1)+2\ln(x+2)+C \right ]$

$=\frac{1}{x-1}+\frac{2}{x+2}+0$

$=\frac{x+2}{(x-1)(x+2)}+\frac{2(x-1)}{(x-1)(x+2)}$

$=\frac{x+2+2(x-1)}{(x-1)(x+2)}$

$=\frac{3x}{x^2+x-2}$

$\begin{align*}&\text{Evaluate the integral: }\&\int_{10}^{15}(\frac{x^{2}}{10})dx\end{align*}$

Explanation

$\begin{align*}&\text{In performing an integral, familiarity with derivative rules}\&\text{can prove useful. After all, an integral is essentially the}\&\text{opposite operation of a derivative. Consider how in a derivative}\&\text{we’d use the chain rule to multiply the derivative of the “inside”}\&\text{of a function by the derivative of the outside? Integrals, in}\&\text{similar fashion, entail a division.}\&\text{Utilizing integral rules:}\&\int x^{a}=\frac{x^{a+1}}{a+1}\&\int_{10}^{15}(\frac{x^{2}}{10})dx=\frac{x^{3}}{30}|_{10}^{15}\&\text{Now we just subtract the function value at the lower bound from that of the upper bound:}\&\int_{10}^{15}(\frac{x^{2}}{10})dx=\frac{15^{3}}{30}-(\frac{10^{3}}{30})\&\int_{10}^{15}(\frac{x^{2}}{10})dx=79.17\end{align*}$