How do you integrate $\int x^{2} e^{\frac{x^{3}}{2}} d x$ $int x^2e^(x^3/2)dx$ from $[- 2, 3]$ $[-2,3]$ ?

Question

How do you integrate $\int x^{2} e^{\frac{x^{3}}{2}} d x$ $int x^2e^(x^3/2)dx$ from $[- 2, 3]$ $[-2,3]$ ?

1 Answer

Impact of this question

2592 views around the world

You can reuse this answer
Creative Commons License

Answer 1 · 2017-01-14T18:44:52

$\frac{2 e^{\frac{35}{2}} - 2}{3 e^{4}}$ $(2e^(35/2)-2)/(3e^4)$

Explanation:

$I = \int_{- 2}^{3} x^{2} e^{\frac{x^{3}}{2}} d x$ $I=int_(-2)^3x^2e^(x^3/2)dx$

Use the substitution:

$u = \frac{x^{3}}{2} \Rightarrow d u = \frac{3}{2} x^{2} d x$ $u=x^3/2" "=>" "du=3/2x^2dx$

We already have $x^{2} d x$ $x^2dx$ in the integral, but are off by a factor of $\frac{3}{2}$ $3/2$ . Also note that changing from $x$ $x$ to $u$ $u$ will change our bounds:

$x = 3 \Rightarrow u = \frac{3^{3}}{2} = \frac{27}{2}$ $x=3" "=>" "u=3^3/2=27/2$

$x = - 2 \Rightarrow u = \frac{{(- 2)}^{3}}{2} = - 4$ $x=-2" "=>" "u=(-2)^3/2=-4$

Thus:

$I = \frac{2}{3} \int e^{\frac{x^{3}}{2}} (\frac{3}{2} x^{2} d x) = \frac{2}{3} \int_{- 4}^{\frac{27}{2}} e^{u} d u$ $I=2/3inte^(x^3/2)(3/2x^2dx)=2/3int_(-4)^(27/2)e^udu$

The integral of $e^{u}$ $e^u$ is itself:

$= \frac{2}{3} {[e^{u}]}_{- 4}^{u} = \frac{2}{3} (e^{\frac{27}{2}} - e^{- 4}) = \frac{2}{3} (\frac{e^{\frac{35}{2}} - 1}{e^{4}}) = \frac{2 e^{\frac{35}{2}} - 2}{3 e^{4}}$ $=2/3[e^u]_(-4)^(27/2)=2/3(e^(27/2)-e^(-4))=2/3((e^(35/2)-1)/e^4)=(2e^(35/2)-2)/(3e^4)$

Science

Math

Humanities

... and beyond

How do you integrate $\int x^{2} e^{\frac{x^{3}}{2}} d x$ $int x^2e^(x^3/2)dx$ from $[- 2, 3]$ $[-2,3]$ ?

1 Answer

Explanation:

Related questions

Impact of this question

Science

Math

Humanities

... and beyond

How do you integrate int x^2e^(x^3/2)dx∫x2ex32dxint x^2e^(x^3/2)dx from [-2,3][−2,3][-2,3]?

1 Answer

Explanation:

Related questions

Impact of this question

How do you integrate $\int x^{2} e^{\frac{x^{3}}{2}} d x$ $int x^2e^(x^3/2)dx$ from $[- 2, 3]$ $[-2,3]$ ?