# Use a table of integrals to evaluate the definite integral. int_0^3 sqrt(x^2+16)dx

smileycellist2 2021-02-05 Answered
Use a table of integrals to evaluate the definite integral.
${\int }_{0}^{3}\sqrt{{x}^{2}+16}dx$
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## Expert Answer

Nathanael Webber
Answered 2021-02-06 Author has 117 answers
Step 1
We have to evaluate the definite integral using table of integrals:
${\int }_{0}^{3}\sqrt{{x}^{2}+16}dx$
Rewriting the integral,
${\int }_{0}^{3}\sqrt{{x}^{2}+16}dx={\int }_{0}^{3}\sqrt{{x}^{2}+{4}^{2}}dx$
We know from table of integrals,
$\int \sqrt{{x}^{2}+{a}^{2}}dx=\frac{x\sqrt{{x}^{2}+{a}^{2}}}{2}+\frac{{a}^{2}}{2}\mathrm{ln}\left(x+\sqrt{{x}^{2}+{a}^{2}}\right)+C$
Where, C is an arbitrary constant.
Step 2
Applying above formula,
$\int \sqrt{{x}^{2}+{a}^{2}}dx=\frac{x\sqrt{{x}^{2}+{a}^{2}}}{2}+\frac{{a}^{2}}{2}\mathrm{ln}\left(x+\sqrt{{x}^{2}+{a}^{2}}\right)+C$
${\int }_{0}^{3}\sqrt{{x}^{2}+{4}^{2}}dx={\left[\frac{x\sqrt{{x}^{2}+{4}^{2}}}{2}+\frac{{4}^{2}}{2}\mathrm{ln}\left(x+\sqrt{{x}^{2}+{4}^{2}}\right)\right]}_{0}^{3}$
$={\left[\frac{x\sqrt{{x}^{2}+16}}{2}+\frac{16}{2}\mathrm{ln}\left(x+\sqrt{{x}^{2}+16}\right)\right]}_{0}^{3}$
$=\left[\frac{3\sqrt{{3}^{2}+16}}{2}+\frac{16}{2}\mathrm{ln}\left(3+\sqrt{{3}^{2}+16}\right)-\left(\left(0\frac{\sqrt{{0}^{2}+16}}{2}\right)+\frac{16}{2}\mathrm{ln}\left(0+\sqrt{{0}^{2}+16}\right)\right)\right]$
$=\left[\frac{x\sqrt{9+16}}{2}+\frac{16}{2}\mathrm{ln}\left(3+\sqrt{9+16}\right)-\frac{3\sqrt{16}}{2}-\frac{16}{2}\mathrm{ln}\left(\sqrt{0+16}\right)\right]$
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