StudySmarter AI is coming soon!

- :00Days
- :00Hours
- :00Mins
- 00Seconds

A new era for learning is coming soonSign up for free

Suggested languages for you:

Americas

Europe

Q. 79

Expert-verifiedFound in: Page 465

Book edition
1st

Author(s)
Peter Kohn, Laura Taalman

Pages
1155 pages

ISBN
9781429241861

Solve given definite integrals.

${\int}_{0}^{4}x\sqrt{{x}^{2}+4}dx$

$\frac{40\sqrt{5}}{3}-\frac{8}{3}\text{.}$

The integral is as follows.

${\int}_{0}^{4}\u200ax\sqrt{{x}^{2}+4}dx$

The objective is to solve the integral.

The integral is solved below.

$\int x\sqrt{{x}^{2}+4}dx$

$=\int \frac{\sqrt{u}}{2}du\phantom{\rule{1em}{0ex}}\left[u={x}^{2}+4,du=2xdx\right]$

$\begin{array}{r}=\frac{1}{2}\int \sqrt{u}du\\ =\frac{{u}^{\frac{3}{2}}}{3}\\ =\frac{{\left({x}^{2}+4\right)}^{\frac{3}{2}}}{3}\\ =\frac{1}{3}{\left({x}^{2}+4\right)}^{\frac{3}{2}}\end{array}$

The definite integral is solved below.

$\begin{array}{r}{\int}_{0}^{4}\u200ax\sqrt{{x}^{2}+4}dx\\ ={\left[\frac{1}{3}{\left({x}^{2}+4\right)}^{\frac{3}{2}}\right]}_{0}^{4}\\ =\left[\frac{1}{3}{\left({4}^{2}+4\right)}^{\frac{3}{2}}\right]-\left[\frac{1}{3}{\left({0}^{2}+4\right)}^{\frac{3}{2}}\right]\\ =\frac{40\sqrt{5}}{3}-\frac{8}{3}\end{array}$

Therefore, the value is $\frac{40\sqrt{5}}{3}-\frac{8}{3}\text{.}$

94% of StudySmarter users get better grades.

Sign up for free