Evaluate the definite integral $\int_0^6 \frac{x}{\sqrt{9+2x^2}}dx$

Question

Evaluate the integral $\int_0^6 \frac{x}{\sqrt{9+2x^2}}dx$

Section 5.4

Can somebody verify this solution for me? Thanks!

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$\int_0^6 \frac{x}{\sqrt{9+2x^2}}dx$

$=\int_0^6 \frac{x}{\sqrt{9+2x^2}}dx$

Let $u=9+2x^2$. Then $\frac{du}{dx}=4x$ and so $\frac{du}{4x}=2x$. Thus we have:

$=\int_0^6 \frac{x}{\sqrt{9+2x^2}}dx$

$=\int_0^6 \frac{x}{\sqrt{u}}\frac{du}{4x}$

$=\frac{1}{4} \int_0^6 u^{\frac{-1}{2}}du$

$=\frac{1}{4} \frac{u^{\frac{1}{2}}}{\frac{1}{2}}|_0^6$

$=\frac{1}{2} u^{\frac{1}{2}}||_0^6$

Make sure you substitute away the dummy variable $u$ before you evaluate at $x=0$ and $x=6$!!

$=\frac{1}{2}(9+2x^2)^{\frac{1}{2}}|_0^6$

$=\frac{1}{2}((9+2(6)^2)^{\frac{1}{2}}-(9+2(0)^2)^{\frac{1}{2}})$

$=\frac{1}{2}(81^{\frac{1}{2}}-9^{\frac{1}{2}})$

$=\frac{1}{2}(9-3)$

$=\frac{1}{2}6$

$=3$

Answer 1

For the integrand $\frac{x}{\sqrt{2 x^2 + 9}}$, substitute $u = 2 x^2 + 9$ and $du = 4xdx$.

This gives a new lower bound $u = 9 + 2 0^2 = 9$ and upper bound:

$u = 9 + 2 6^2 = 81 \Longrightarrow = \frac{1}{4} \int_{9}^{81} \frac{1}{\sqrt{u}}{du}$

Apply the fundamental theorem of calculus.

The antiderivative of $\frac{1}{\sqrt(u)}$ is $2 \sqrt{u}: = \frac{1}{2}\sqrt{u} \ \mid_{9}^{81}$

Evaluate the antiderivative at the limits and subtract.

$\frac{1}{2}{\sqrt(u)}\mid _{9}^{81} = \frac{\sqrt(81)}{2} - \frac{\sqrt(9)}{2} = 3$

Hence your answer is correct.