Expert answers to "Use a table of integrals to evaluate the..."

smileycellist2

Answered question

2021-02-05

Use a table of integrals to evaluate the definite integral.

\int_{0}^{3} \sqrt{x^{2} + 16} d x

Nathanael Webber

Skilled2021-02-06Added 117 answers

Step 1
We have to evaluate the definite integral using table of integrals:

\int_{0}^{3} \sqrt{x^{2} + 16} d x

Rewriting the integral,

\int_{0}^{3} \sqrt{x^{2} + 16} d x = \int_{0}^{3} \sqrt{x^{2} + 4^{2}} d x

We know from table of integrals,

\int \sqrt{x^{2} + a^{2}} d x = \frac{x \sqrt{x^{2} + a^{2}}}{2} + \frac{a^{2}}{2} \ln (x + \sqrt{x^{2} + a^{2}}) + C

Where, C is an arbitrary constant.
Step 2
Applying above formula,

\int \sqrt{x^{2} + a^{2}} d x = \frac{x \sqrt{x^{2} + a^{2}}}{2} + \frac{a^{2}}{2} \ln (x + \sqrt{x^{2} + a^{2}}) + C

\int_{0}^{3} \sqrt{x^{2} + 4^{2}} d x = {[\frac{x \sqrt{x^{2} + 4^{2}}}{2} + \frac{4^{2}}{2} \ln (x + \sqrt{x^{2} + 4^{2}})]}_{0}^{3}

= {[\frac{x \sqrt{x^{2} + 16}}{2} + \frac{16}{2} \ln (x + \sqrt{x^{2} + 16})]}_{0}^{3}

= [\frac{3 \sqrt{3^{2} + 16}}{2} + \frac{16}{2} \ln (3 + \sqrt{3^{2} + 16}) - ((0 \frac{\sqrt{0^{2} + 16}}{2}) + \frac{16}{2} \ln (0 + \sqrt{0^{2} + 16}))]

= [\frac{x \sqrt{9 + 16}}{2} + \frac{16}{2} \ln (3 + \sqrt{9 + 16}) - \frac{3 \sqrt{16}}{2} - \frac{16}{2} \ln (\sqrt{0 + 16})]

= [\frac{3 \sqrt{25}}{2} + 8 \ln (3 + \sqrt{25}) -

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