$$\int_0^1 xe^{\sqrt{x}} dx = ? $$
All I can think of is the integration by parts rule, where $ u = x $ and $ dv= e^{\sqrt(x)} $ $ \Rightarrow du = 1$ and $ v= e^{\sqrt(x)} $
The answer I get is $e^{\sqrt(x)}(x-1)$ , which is wrong.
Can anyone please explain in detail?
Let $u=\sqrt{x}$ so $du=\frac{1}{2\sqrt{x}}dx=\frac{1}{2u}dx$ so the integral becomes
$$ \int_0^1 2u^3e^u du$$
then evaluate this by integrating by parts.
You can do integration by parts like this, without substituting. Of course, substituting is fine and all, but you'll have to use Integration by Parts three times either way.
$$\begin{align} \int_0^1xe^{\sqrt{x}}\,dx &=\int_0^1\frac{2x\sqrt{x}}{2\sqrt{x}}e^{\sqrt{x}}\,dx\\ &=\int_0^12x\sqrt{x}\left(\frac{1}{2\sqrt{x}}e^{\sqrt{x}}\,dx\right)\\ &=\left[2x\sqrt{x}e^{\sqrt{x}}\right]_0^1-\int_0^13\sqrt{x}\,e^{\sqrt{x}}\,dx\\ &=\left[2x\sqrt{x}e^{\sqrt{x}}\right]_0^1-\int_0^13\sqrt{x}\,e^{\sqrt{x}}\,dx \end{align}$$
And we've reduced the intebrand from $cx^1e^{\sqrt{x}}$ to $cx^{1/2}e^{\sqrt{x}}$. Repeat the technique two more times to bring it to $cx^{-1/2}e^{\sqrt{x}}$, and then you can just directly antidifferentiate.
In fact, $$ \int xe^{\sqrt{x}}dx = 2\int x\sqrt{x}\dfrac{e^{\sqrt{x}}}{2\sqrt{x}}dx = 2\int x\sqrt{x} e^{\sqrt{x}}d(\sqrt{x}) = 2\int u^{3}e^u du = \dfrac{2}{D}(u^3e^u) $$ $$ = 2e^u \dfrac{1}{1 + D}u^3 = 2e^u(1 - D + D^2 - D^3)u^3 = 2e^u[u^3 - 3u^2 + 6u - 6] $$ $$ =2e^{\sqrt{x}}[x^{3/2} - 3x + 6\sqrt{x} - 6] + C $$
