Integral of $\int ^{5}_{1}\int ^{\sqrt{x-1}}_{y=0}{ye^{(x-1)^2}}dydx$

Question

$\int ^{5}_{1}\int ^{\sqrt{x-1}}_{y=0}{ye^{(x-1)^2}}dydx$

First, I changed the integration from $x$ to $y$.

However, I get $\int ^{5}_{1}\dfrac{\left( x-1\right) }{2}e^{\left( x-1\right) ^{2}}dx$ and I get $\int \dfrac{1}{4}\sqrt{u}e^{u}du$.

I can't go from here. How to integrate this?

Answer 1

You are making the substitution wrong. If $u=(x-1)^2$, then $du=2(x-1)dx$, and the factor $(x-1)/2$ becomes simply $1/4$, not $\frac14\sqrt u$.

Answer 2

First integrate the function with respect to $y$:

$$\int ^{5}_{1}\int ^{\sqrt{x-1}}_{y=0}{ye^{(x-1)^2}}dydx = \left[\int ^{5}_{1}{\frac{y^2}{2}e^{(x-1)^2}}dx\right]^{\sqrt{x-1}}_{y=0} \\ = \int ^{5}_{1}{\frac{(x-1)}{2}e^{(x-1)^2}}dx$$

Now, suppose $u = (x-1)^2$. Defferntiating gives you $\frac{du}{dx} = 2(x-1)$ or $du = 2(x-1)dx$. The limits are: When $x = 1$, $u = 0$ an when $x = 5$, $u = 4^2 = 16$. Nowe, apply $u = (x-1)^2$, $(x-1)dx = \frac12 du$, and limits in the equation:

$$\int ^{5}_{1}{\frac{(x-1)}{2}e^{(x-1)^2}}dx = \int ^{16}_{0}\frac{1}{4}e^{u}du \\ = \left[\frac{1}{4}e^{u}\right]^{16}_{0} = \frac{1}{4}(e^{16} - e^{0}) = \frac{1}{4}(e^{16} - 1)$$