Question

Show that d/dx (x^2 e^(x^2 ) )=2xe^(x^2 )+2x^3 e^(x^2 ). Use this result to evaluate ∫_0^1▒x^3 e^(x^2 ) dx

Answer 1

I am doing the Integration..trying

Answer 2

use the product rule...
d/dx ( f(x)g(x)) = f '(x)g(x)+f(x)g'(x)
d/dx (x^2e^(x^2))= 2xe^(x^2)+2x^3e^(x^2)

Answer 3

\[\int_0^1 x^3e^{x^2}\]This it it Right ?

Answer 4

Yes

Answer 5

\[Now\:\:\:I\:\:\:say\:\:t = x^4\]
\[dx = \frac{dt}{4x^3}\]

Answer 6

or lets do \[x^2 = t\]
\[dx = \frac{dt}[2x}\]
\[\frac{1}{2}\int_0^1t* e^t\]By Parts
\[ \frac{te^t - e^t}{2} |_0^1\]
\[ 0 - \frac{1}{2}\]
Hence \[-\frac{1}{2}\]

Answer 7

the ans is 1/2 not -1/2

Answer 8

oh yeah -(-1/2) sorry my mistake

Answer 9

Using the result of differentiation you rearrange the equation to get
x^3*e^(x^2)=-x*e^(x^2)+ 0.5 d/dx(x^2*e^(x^2))
Now integrate both sides from 0 to 1 and use the fundamental theorem of calculus to get

integral 0 to 1 of x^3*e^(x^2) [This is what we want to solve for]
=integral -x*e^(x^2)dx +.5*x^2*e^(x^2)/2|evaluated at 1 and 0
The integral can be done by substitution t=x^2 and that equals -e/2
The second is just evaluating at 1 and zero and getting 1/2 +e/2
Thus together they add up to 1/2

This method relies more on the differentiation that we did earlier.