Question

question 16 · 1 point
evaluate the definite integral below.
(enter an exact answer.)
provide your answer below:
\\(\int_{2}^{3}3x\sqrt{3x^{2}-7}dx = \square\\)

Explanation:

Step1: Use substitution

Let $u = 3x^{2}-7$, then $du=6x dx$, and $x dx=\frac{1}{6}du$. When $x = 2$, $u=3\times2^{2}-7=12 - 7 = 5$. When $x = 3$, $u=3\times3^{2}-7=27 - 7 = 20$.

Step2: Rewrite the integral

The integral $\int_{2}^{3}3x\sqrt{3x^{2}-7}dx=\frac{3}{6}\int_{5}^{20}\sqrt{u}du=\frac{1}{2}\int_{5}^{20}u^{\frac{1}{2}}du$.

Step3: Integrate $u^{\frac{1}{2}}$

Using the power - rule for integration $\int x^{n}dx=\frac{x^{n + 1}}{n+1}+C$ ($n
eq - 1$), we have $\frac{1}{2}\times\frac{u^{\frac{1}{2}+1}}{\frac{1}{2}+1}\big|_{5}^{20}=\frac{1}{2}\times\frac{u^{\frac{3}{2}}}{\frac{3}{2}}\big|_{5}^{20}=\frac{1}{3}u^{\frac{3}{2}}\big|_{5}^{20}$.

Step4: Evaluate the definite integral

$\frac{1}{3}(20^{\frac{3}{2}}-5^{\frac{3}{2}})=\frac{1}{3}(20\sqrt{20}-5\sqrt{5})=\frac{1}{3}(40\sqrt{5}-5\sqrt{5})=\frac{35\sqrt{5}}{3}$.

Answer:

$\frac{35\sqrt{5}}{3}$