(1) 11. Find the area between $$ y=(x-2) e^{x} $$ and the $$ x $$-axis from $$ x=2 $$ to $$ x=4 $$. (1) 12. Find the area between $$ y=(x+1) \ln x $$ and the $$ x $$-axis from $$ x=1 $$ to $$ x=e $$. (1) Exercises $$ 13-22 $$ are mixed-some require integration by parts, while others can be integrated by using techniques discussed in the chapter on Integration. $$ \begin{array}{ll} ext { 13. } \int x^{2} e^{2 x} d x & ext { 14. } \int \frac{x^{2} d x}{2 x^{3}+1} \ ext { 15. } \int x^{2} \sqrt{x+4} d x & ext { 16. } \int(2 x-1) \ln (3 x) d x \ ext { 17. } \int(8 x+10) \ln (5 x) d x & ext { 18. } \int x^{3} e^{x^{4}} d x\end{array} $$

Question

(1) 11. Find the area between $$ y=(x-2) e^{x} $$ and the $$ x $$-axis from $$ x=2 $$ to $$ x=4 $$. (1) 12. Find the area between $$ y=(x+1) \ln x $$ and the $$ x $$-axis from $$ x=1 $$ to $$ x=e $$. (1) Exercises $$ 13-22 $$ are mixed-some require integration by parts, while others can be integrated by using techniques discussed in the chapter on Integration. $$ \begin{array}{ll}	ext { 13. } \int x^{2} e^{2 x} d x & 	ext { 14. } \int \frac{x^{2} d x}{2 x^{3}+1} \ 	ext { 15. } \int x^{2} \sqrt{x+4} d x & 	ext { 16. } \int(2 x-1) \ln (3 x) d x \ 	ext { 17. } \int(8 x+10) \ln (5 x) d x & 	ext { 18. } \int x^{3} e^{x^{4}} d x\end{array} $$

UpStudy AI · Accepted Answer

To find the area between the given functions and the x-axis, we need to integrate the functions with respect to x and then subtract the area under the x-axis.

Given functions:
1. $$ y = (x-2) e^{x} $$ from $$ x = 2 $$ to $$ x = 4 $$
2. $$ y = (x+1) \ln x $$ from $$ x = 1 $$ to $$ x = e $$

Let's calculate the area for each function:

1. For $$ y = (x-2) e^{x} $$ from $$ x = 2 $$ to $$ x = 4 $$:
   The area under the curve is given by the integral of $$ y $$ with respect to x from 2 to 4:
   $$ \text{Area} = \int_{2}^{4} (x-2) e^{x} dx $$

2. For $$ y = (x+1) \ln x $$ from $$ x = 1 $$ to $$ x = e $$:
   The area under the curve is given by the integral of $$ y $$ with respect to x from 1 to e:
   $$ \text{Area} = \int_{1}^{e} (x+1) \ln x dx $$

Now, let's calculate the integrals using integration by parts.
Evaluate the integral by following steps:
- step0: Evaluate using partial integration formula:
  $$\int_{2}^{4} \left(x-2\right)e^{x} dx$$
- step1: Evaluate the integral:
  $$\int \left(x-2\right)e^{x} dx$$
- step2: Prepare for integration by parts:
  $$\begin{align}&u=x-2\&dv=e^{x}dx\end{align}$$
- step3: Calculate the derivative:
  $$\begin{align}&du=dx\&dv=e^{x}dx\end{align}$$
- step4: Evaluate the integral:
  $$\begin{align}&du=dx\&v=e^{x}\end{align}$$
- step5: Substitute the values into formula:
  $$\left(x-2\right)e^{x}-\int 1\times e^{x} dx$$
- step6: Calculate:
  $$\left(x-2\right)e^{x}-\int e^{x} dx$$
- step7: Evaluate the integral:
  $$\left(x-2\right)e^{x}-e^{x}$$
- step8: Expand the expression:
  $$xe^{x}-2e^{x}-e^{x}$$
- step9: Subtract the terms:
  $$xe^{x}-3e^{x}$$
- step10: Return the limits:
  $$\left(xe^{x}-3e^{x}\right)\bigg |_{2}^{4}$$
- step11: Calculate the value:
  $$e^{4}+e^{2}$$
Calculate the integral $$ \int_{1}^{e} (x+1) \ln x dx $$.
Evaluate the integral by following steps:
- step0: Evaluate using partial integration formula:
  $$\int_{1}^{e} \left(x+1\right)\times \ln{\left(x\right)} dx$$
- step1: Evaluate the integral:
  $$\int \left(x+1\right)\times \ln{\left(x\right)} dx$$
- step2: Prepare for integration by parts:
  $$\begin{align}&u=\ln{\left(x\right)}\&dv=\left(x+1\right)dx\end{align}$$
- step3: Calculate the derivative:
  $$\begin{align}&du=\frac{1}{x}dx\&dv=\left(x+1\right)dx\end{align}$$
- step4: Evaluate the integral:
  $$\begin{align}&du=\frac{1}{x}dx\&v=\frac{x^{2}}{2}+x\end{align}$$
- step5: Substitute the values into formula:
  $$\ln{\left(x\right)}\times \left(\frac{x^{2}}{2}+x\right)-\int \frac{1}{x}\times \left(\frac{x^{2}}{2}+x\right) dx$$
- step6: Calculate:
  $$\frac{x^{2}\ln{\left(x\right)}}{2}+x\ln{\left(x\right)}-\int \frac{x+2}{2} dx$$
- step7: Evaluate the integral:
  $$\frac{x^{2}\ln{\left(x\right)}}{2}+x\ln{\left(x\right)}-\frac{x^{2}}{4}-x$$
- step8: Simplify the expression:
  $$\frac{1}{2}x^{2}\ln{\left(x\right)}+x\ln{\left(x\right)}-\frac{x^{2}}{4}-x$$
- step9: Return the limits:
  $$\left(\frac{1}{2}x^{2}\ln{\left(x\right)}+x\ln{\left(x\right)}-\frac{x^{2}}{4}-x\right)\bigg |_{1}^{e}$$
- step10: Calculate the value:
  $$\frac{e^{2}+5}{4}$$
The area between the function $$ y = (x-2) e^{x} $$ and the x-axis from $$ x = 2 $$ to $$ x = 4 $$ is $$ e^{4} + e^{2} $$.

The area between the function $$ y = (x+1) \ln x $$ and the x-axis from $$ x = 1 $$ to $$ x = e $$ is $$ \frac{e^{2}+5}{4} $$.
The area between $$ y=(x-2) e^{x} $$ and the x-axis from $$ x=2 $$ to $$ x=4 $$ is $$ e^{4} + e^{2} $$. The area between $$ y=(x+1) \ln x $$ and the x-axis from $$ x=1 $$ to $$ x=e $$ is $$ \frac{e^{2}+5}{4} $$.

(1) 11. Find the area between and the -axis from
to .
(1) 12. Find the area between and the -axis from
to .
(1) Exercises are mixed-some require integration by parts,
while others can be integrated by using techniques discussed in
the chapter on Integration.

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(1) 11. Find the area between and the -axis from to . (1) 12. Find the area between and the -axis from to . (1) Exercises are mixed-some require integration by parts, while others can be integrated by using techniques discussed in the chapter on Integration.