1. ## A quick check on several integral problems,please

These problems are part of my homework. However, I am bit unsure about some of them, so I'd like to ask you whether my answers are correct. If you find mistakes and have the time please write me the full solution. Thx

Pr.1 $\displaystyle \int xe^x dx$ Let $u=x, \frac{du}{dx}=1, du=dx$ $\displaystyle, Let v= \int e^x=e^x
$
$\displaystyle\int udv=uv-\int vdu = xe^x-\int e^x dx=xe^x-e^x=e^x(x-1)$

Pr.2 $\displaystyle \int x^2 e^{x^{3}}, Let u=x^2,\frac{du}{dx}=2x, du=2x dx$
Let $\displaystylev=\int e^{x^{3}}= e^{x^{2}}/x$
$\displaystyle\int u dv= x^2e^{x^{2}}/x- \int 2x{e^{x^{2}}/x=xe^{x^{2}}-2e^{x^{2}}=e^{x^{2}}(x-2)$

Pr.3 $\displaystyle\int x^2 e^x dx, Let u=x^2,du/dx=2x, du= 2x dx$ $\displaystyleLet v=\int e^x=e^x$ Then $\int u dv=x^2 e^x -\int e^x2x dx$ $\displaystyle,Let u_{2}=e^x,du_{2}/dx=e^x, du_{2}=e^xdx$ $\displaystyleLet v_{2}=\int 2x=2,\int u_{2}dv_{2}=e^x2-\int 2e^x dx=e^x2-2e^x=0$

$\displaystyle\int u dv=x^2e^x$

Pr.4 $\displaystyle\int x e^{x^{2}}dx$ Let $\displaystyleu=x,du/dx=1,du=dx, Let v=\int e^{x^{2}}=e^{x^{2}}$
$\displaystyle\int u dv=xe^{x^{2}}-\int e^{x^{2}} dx =xe^{x^{2}}-e^{x^{2}}=e^{x^{2}}(x-1)$

2. 1 correct

2 incorrect

3 not sure what your final answer is here but you don't have it correct

4 incorrect

You can always take the derivative of your solutions as a check too.

3. Originally Posted by Boz
These problems are part of my homework. However, I am bit unsure about some of them, so I'd like to ask you whether my answers are correct. If you find mistakes and have the time please write me the full solution. Thx

Pr.2 $\displaystyle \int x^2 e^{x^{3}}, Let u=x^2,\frac{du}{dx}=2x, du=2x dx$
Let $\displaystylev=\int e^{x^{3}}= e^{x^{2}}/x$
$\displaystyle\int u dv= x^2e^{x^{2}}/x- \int 2x{e^{x^{2}}/x=xe^{x^{2}}-2e^{x^{2}}=e^{x^{2}}(x-2)$

Pr.4 $\displaystyle\int x e^{x^{2}}dx$ Let $\displaystyleu=x,du/dx=1,du=dx, Let v=\int e^{x^{2}}=e^{x^{2}}$
$\displaystyle\int u dv=xe^{x^{2}}-\int e^{x^{2}} dx =xe^{x^{2}}-e^{x^{2}}=e^{x^{2}}(x-1)$
For No. 2.. Use the Substitution Rule..

Let $u=x^3$ and $du=3x^2\;dx \implies dx = \dfrac{du}{3x^2}$

so, $\displaystyle \int x^2 e^{x^{3}} = \int x^2\;e^u\;\dfrac{du}{3x^2} = \dfrac{1}{3}\int e^u\;du$

integrate and plug back u..

Do the same for No. 4. Let $u=x^2$

What are you doing for No 3? You need integration by parts!

4. Prb3:
$\displaystyle u=x^2$
$\displaystyle v=e^x$
$\displaystyle du=2xdx$
$\displaystyle dv=e^xdx$

$\displaystyle \int udv=uv-\int vdu$
$\displaystyle \int x^2 e^x dx=x^2e^x-2\int e^xxdx$

Then int by parts again to get the answer.

5. Originally Posted by integral
Prb3:
$\displaystyle u=x^2$
$\displaystyle v=e^x$
$\displaystyle du=2xdx$
$\displaystyle dv=e^xdx$

$\displaystyle \int udv=uv-\int vdu$
$\displaystyle \int x^2 e^x dx=x^2e^x-2\int e^xxdx$

Then int by parts again to get the answer.
So, the answer would be $e^x(x^2-2x+4)$??

pr.2 $\frac{e{x^3}}{3}$?
pr.4 $\frac{e{x^2}}{2}$?

6. Originally Posted by integral
Prb3:
$\displaystyle u=x^2$
$\displaystyle v=e^x$
$\displaystyle du=2xdx$
$\displaystyle dv=e^xdx$

$\displaystyle \int udv=uv-\int vdu$
$\displaystyle \int x^2 e^x dx=x^2e^x-2\int e^xxdx$

Then int by parts again to get the answer.
So, the answer would be $e^x(x^2-2x+4)$??

pr.2 $\frac{e^{x^3}}{3}$?
pr.4 $\frac{e^{x^2}}{2}$?

7. Originally Posted by Boz
So, the answer would be $e^x(x^2-2x+4)$ I think you made a calculation mistake. It should be $e^x(x^2-2x+2)$??

pr.2 $\frac{e^{x^3}}{3}$?
pr.4 $\frac{e^{x^2}}{2}$?