Computing the definite integral $\int _0^a \:x \sqrt{x^2+a^2} \,\mathrm d x$

Question

Compute the following definite integral $$\int _0^a \:x \sqrt{x^2+a^2} \,\mathrm d x$$

This is what I did:

$u = x^2 + a^2 $

$du/dx = 2x$

$du = 2xdx$

$1/2 du = x dx$

$\int _0^a\:\frac{1}{2}\sqrt{u}du = \frac{1}{2}\cdot \frac{u^{\frac{3}{2}}}{\left(\frac{3}{2}\right)}$ from $0$ to $a$.

$\frac{1}{3}\cdot \left(x^2+a^2\right)^{\frac{3}{2}}$ from $0$ to $a$.

I eventually got:

$\frac{1}{3}\left(81+a^2\right)^{\frac{3}{2}}-\frac{1}{3}\left(a^2\right)^{\frac{3}{2}}$

but this was incorrect.

The correct answer was:

$\frac{1}{3}\left(2\sqrt{2}-1\right)a^3$

Any help?

Answer 1

Hint

You made mistakes here:

$$\frac{1}{3}\cdot \left(x^2+a^2\right)^{\frac{3}{2}}\big |_0^a=\frac 1 3 (2a^2)^{\frac 3 2}-\frac {a^3}{3}=\frac 1 3 (\sqrt{2^3}|a|^3)-\frac {a^3}{3}=....$$

You were almost done...

Answer 2

You calculated the antiderivative correctly.

$$\frac{1}{3}\cdot \left(x^2+a^2\right)^{\frac{3}{2}}$$ from $0$ to $a$ is $$\frac{1}{3}\cdot \left(a^2+a^2\right)^{\frac{3}{2}}-\frac{1}{3}\cdot \left(0^2+a^2\right)^{\frac{3}{2}}=\\\frac{1}{3}\cdot((2a^2)^\frac{3}{2}-(a^2)^\frac{3}{2})=\frac{1}{3}\cdot(2^\frac{3}{2}a^3-a^3)=\frac{1}{3}(2^\frac{3}{2}-1)a^3$$

Which explains the answer in your book. I don't know how you got $$\frac{1}{3}\left(81+a^2\right)^{\frac{3}{2}}-\frac{1}{3}\left(a^2\right)^{\frac{3}{2}}$$ from the previous step.

Answer 3

You can simply use the reverse chain rule for integration and you get: $$\int_0^a x\sqrt{x^2+a^2}dx=\frac 12\int_0^a 2x\sqrt{x^2+a^2}dx=\left[\frac{(x^2+a^2)^\frac 32}3\right]_0^a=\\=\frac{(2a^2)^\frac 32}3-\frac{(a^2)^\frac 32}3=\color{red}{\frac{(2\sqrt 2-1)}3|a|^3}$$