Math / Calculus

Question\begin{tabular}{|l|} \hline Score \\ \hline \end{tabular}
Part V Evaluate the following integral. (5 scores per question. The total is 10 scores.)
18. $\int_{-1}^{1}\left(e^{2 x}-\frac{1}{2+x}\right) d x$ .

Studdy Solution

STEP 1

Assumptions
1. We need to evaluate the definite integral: $\int_{-1}^{1}\left(e^{2x} - \frac{1}{2+x}\right) dx$ .
2. The integral is over the interval from $x = -1$ to $x = 1$ .
3. The function $e^{2x}$ is continuous over the interval $[-1, 1]$ .
4. The function $\frac{1}{2+x}$ is continuous over the interval $[-1, 1]$ , except at $x = -2$ , which is outside the interval.
5. We will evaluate the integral by splitting it into two separate integrals.

STEP 2

Split the integral into two separate integrals:
$\int_{-1}^{1}\left(e^{2x} - \frac{1}{2+x}\right) dx = \int_{-1}^{1} e^{2x} \, dx - \int_{-1}^{1} \frac{1}{2+x} \, dx$

STEP 3

Evaluate the first integral: $\int_{-1}^{1} e^{2x} \, dx$ .
To do this, use the substitution $u = 2x$ , which gives $du = 2 \, dx$ or $dx = \frac{1}{2} \, du$ .

STEP 4

Change the limits of integration for the substitution:
When $x = -1$ , $u = 2(-1) = -2$ .
When $x = 1$ , $u = 2(1) = 2$ .
Thus, the integral becomes:
$\int_{-2}^{2} e^{u} \cdot \frac{1}{2} \, du = \frac{1}{2} \int_{-2}^{2} e^{u} \, du$

STEP 5

Evaluate the integral $\frac{1}{2} \int_{-2}^{2} e^{u} \, du$ :
The antiderivative of $e^{u}$ is $e^{u}$ , so:
$\frac{1}{2} \left[ e^{u} \right]_{-2}^{2} = \frac{1}{2} \left( e^{2} - e^{-2} \right)$

STEP 6

Evaluate the second integral: $\int_{-1}^{1} \frac{1}{2+x} \, dx$ .
Use the substitution $v = 2 + x$ , which gives $dv = dx$ .

STEP 7

Change the limits of integration for the substitution:
When $x = -1$ , $v = 2 + (-1) = 1$ .
When $x = 1$ , $v = 2 + 1 = 3$ .
Thus, the integral becomes:
$\int_{1}^{3} \frac{1}{v} \, dv$

STEP 8

Evaluate the integral $\int_{1}^{3} \frac{1}{v} \, dv$ :
The antiderivative of $\frac{1}{v}$ is $\ln |v|$ , so:
$\left[ \ln |v| \right]_{1}^{3} = \ln |3| - \ln |1| = \ln 3 - \ln 1$
Since $\ln 1 = 0$ , this simplifies to:
$\ln 3$

STEP 9

Combine the results from STEP_5 and STEP_8 to find the value of the original integral:
$\int_{-1}^{1}\left(e^{2x} - \frac{1}{2+x}\right) dx = \frac{1}{2} (e^{2} - e^{-2}) - \ln 3$

STEP 10

Calculate the final result:
The integral evaluates to:
$\frac{1}{2} (e^{2} - e^{-2}) - \ln 3$
This is the final answer.

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Question\begin{tabular}{|l|} \hline Score \\ \hline \end{tabular}
Part V Evaluate the following integral. (5 scores per question. The total is 10 scores.)
18. $\int_{-1}^{1}\left(e^{2 x}-\frac{1}{2+x}\right) d x$ .

Studdy Solution

STEP 1

STEP 2

Split the integral into two separate integrals:
$\int_{-1}^{1}\left(e^{2x} - \frac{1}{2+x}\right) dx = \int_{-1}^{1} e^{2x} \, dx - \int_{-1}^{1} \frac{1}{2+x} \, dx$

STEP 3

Evaluate the first integral: $\int_{-1}^{1} e^{2x} \, dx$ .
To do this, use the substitution $u = 2x$ , which gives $du = 2 \, dx$ or $dx = \frac{1}{2} \, du$ .

STEP 4

Change the limits of integration for the substitution:
When $x = -1$ , $u = 2(-1) = -2$ .
When $x = 1$ , $u = 2(1) = 2$ .
Thus, the integral becomes:
$\int_{-2}^{2} e^{u} \cdot \frac{1}{2} \, du = \frac{1}{2} \int_{-2}^{2} e^{u} \, du$

STEP 5

Evaluate the integral $\frac{1}{2} \int_{-2}^{2} e^{u} \, du$ :
The antiderivative of $e^{u}$ is $e^{u}$ , so:
$\frac{1}{2} \left[ e^{u} \right]_{-2}^{2} = \frac{1}{2} \left( e^{2} - e^{-2} \right)$

STEP 6

Evaluate the second integral: $\int_{-1}^{1} \frac{1}{2+x} \, dx$ .
Use the substitution $v = 2 + x$ , which gives $dv = dx$ .

STEP 7

Change the limits of integration for the substitution:
When $x = -1$ , $v = 2 + (-1) = 1$ .
When $x = 1$ , $v = 2 + 1 = 3$ .
Thus, the integral becomes:
$\int_{1}^{3} \frac{1}{v} \, dv$

STEP 8

Evaluate the integral $\int_{1}^{3} \frac{1}{v} \, dv$ :
The antiderivative of $\frac{1}{v}$ is $\ln |v|$ , so:
$\left[ \ln |v| \right]_{1}^{3} = \ln |3| - \ln |1| = \ln 3 - \ln 1$
Since $\ln 1 = 0$ , this simplifies to:
$\ln 3$

STEP 9

Combine the results from STEP_5 and STEP_8 to find the value of the original integral:
$\int_{-1}^{1}\left(e^{2x} - \frac{1}{2+x}\right) dx = \frac{1}{2} (e^{2} - e^{-2}) - \ln 3$

STEP 10

Calculate the final result:
The integral evaluates to:
$\frac{1}{2} (e^{2} - e^{-2}) - \ln 3$
This is the final answer.

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Question\begin{tabular}{|l|} \hline Score \\ \hline \end{tabular}Part V Evaluate the following integral. (5 scores per question. The total is 10 scores.)18. ∫−11(e2x−12+x)dx\int_{-1}^{1}\left(e^{2 x}-\frac{1}{2+x}\right) d x∫−11​(e2x−2+x1​)dx.

Studdy Solution

STEP 1

STEP 2

Split the integral into two separate integrals:∫−11(e2x−12+x)dx=∫−11e2x dx−∫−1112+x dx\int_{-1}^{1}\left(e^{2x} - \frac{1}{2+x}\right) dx = \int_{-1}^{1} e^{2x} \, dx - \int_{-1}^{1} \frac{1}{2+x} \, dx∫−11​(e2x−2+x1​)dx=∫−11​e2xdx−∫−11​2+x1​dx

STEP 3

Evaluate the first integral: ∫−11e2x dx\int_{-1}^{1} e^{2x} \, dx∫−11​e2xdx.To do this, use the substitution u=2xu = 2xu=2x, which gives du=2 dxdu = 2 \, dxdu=2dx or dx=12 dudx = \frac{1}{2} \, dudx=21​du.

STEP 4

STEP 5

Evaluate the integral 12∫−22eu du\frac{1}{2} \int_{-2}^{2} e^{u} \, du21​∫−22​eudu:The antiderivative of eue^{u}eu is eue^{u}eu, so:12[eu]−22=12(e2−e−2)\frac{1}{2} \left[ e^{u} \right]_{-2}^{2} = \frac{1}{2} \left( e^{2} - e^{-2} \right)21​[eu]−22​=21​(e2−e−2)

STEP 6

Evaluate the second integral: ∫−1112+x dx\int_{-1}^{1} \frac{1}{2+x} \, dx∫−11​2+x1​dx.Use the substitution v=2+xv = 2 + xv=2+x, which gives dv=dxdv = dxdv=dx.

STEP 7

Change the limits of integration for the substitution:When x=−1x = -1x=−1, v=2+(−1)=1v = 2 + (-1) = 1v=2+(−1)=1.When x=1x = 1x=1, v=2+1=3v = 2 + 1 = 3v=2+1=3.Thus, the integral becomes:∫131v dv\int_{1}^{3} \frac{1}{v} \, dv∫13​v1​dv

STEP 8

STEP 9

Combine the results from STEP_5 and STEP_8 to find the value of the original integral:∫−11(e2x−12+x)dx=12(e2−e−2)−ln⁡3\int_{-1}^{1}\left(e^{2x} - \frac{1}{2+x}\right) dx = \frac{1}{2} (e^{2} - e^{-2}) - \ln 3∫−11​(e2x−2+x1​)dx=21​(e2−e−2)−ln3

STEP 10

Calculate the final result:The integral evaluates to:12(e2−e−2)−ln⁡3\frac{1}{2} (e^{2} - e^{-2}) - \ln 321​(e2−e−2)−ln3This is the final answer.

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Question\begin{tabular}{|l|} \hline Score \\ \hline \end{tabular}
Part V Evaluate the following integral. (5 scores per question. The total is 10 scores.)
18. $\int_{-1}^{1}\left(e^{2 x}-\frac{1}{2+x}\right) d x$ .

Split the integral into two separate integrals:
$\int_{-1}^{1}\left(e^{2x} - \frac{1}{2+x}\right) dx = \int_{-1}^{1} e^{2x} \, dx - \int_{-1}^{1} \frac{1}{2+x} \, dx$

Evaluate the first integral: $\int_{-1}^{1} e^{2x} \, dx$ .
To do this, use the substitution $u = 2x$ , which gives $du = 2 \, dx$ or $dx = \frac{1}{2} \, du$ .

Evaluate the integral $\frac{1}{2} \int_{-2}^{2} e^{u} \, du$ :
The antiderivative of $e^{u}$ is $e^{u}$ , so:
$\frac{1}{2} \left[ e^{u} \right]_{-2}^{2} = \frac{1}{2} \left( e^{2} - e^{-2} \right)$

Evaluate the second integral: $\int_{-1}^{1} \frac{1}{2+x} \, dx$ .
Use the substitution $v = 2 + x$ , which gives $dv = dx$ .

Change the limits of integration for the substitution:
When $x = -1$ , $v = 2 + (-1) = 1$ .
When $x = 1$ , $v = 2 + 1 = 3$ .
Thus, the integral becomes:
$\int_{1}^{3} \frac{1}{v} \, dv$

Combine the results from STEP_5 and STEP_8 to find the value of the original integral:
$\int_{-1}^{1}\left(e^{2x} - \frac{1}{2+x}\right) dx = \frac{1}{2} (e^{2} - e^{-2}) - \ln 3$

Calculate the final result:
The integral evaluates to:
$\frac{1}{2} (e^{2} - e^{-2}) - \ln 3$
This is the final answer.