Mary Buchanan

Answered

2021-12-19

Evaluate the integral by interpreting it in terms of areas. integral:

${\int}_{-3}^{0}(1+\sqrt{9-{x}^{2}})dx$

Answer & Explanation

lenkiklisg7

Expert

2021-12-20Added 29 answers

Step 1

First we break the integral into two parts:

${\int}_{-3}^{0}(1+\sqrt{9-{x}^{2}})dx={\int}_{-3}^{0}1dx+{\int}_{-3}^{0}\sqrt{9-{x}^{2}}dx$

The first term${\int}_{-3}^{0}1dx$ represents area of rectangle with base $b=3$ and height $h=1$

${\int}_{-3}^{0}1dx=b\times h=3$

${\int}_{-3}^{0}\sqrt{9-{x}^{2}}dx$ represent area of quarter circle with radius $r=3$

$\int}_{-3}^{0}\sqrt{9-{x}^{2}}dx=\frac{1}{4}\pi \times {3}^{2}=\frac{9\pi}{4$

${\int}_{-3}^{0}(1+\sqrt{9-{x}^{2}})dx={\int}_{-3}^{0}1dx+{\int}_{-3}^{0}\sqrt{9-{x}^{2}}dx$

$\int}_{-3}^{0}(1+\sqrt{9-{x}^{2}})dx=3+\frac{9\pi}{4$

First we break the integral into two parts:

The first term

Lakisha Archer

Expert

2021-12-21Added 39 answers

Step 1

The given integral is

$I={\int}_{-3}^{0}(1+\sqrt{9-{x}^{x}})dx$

$I={\int}_{-3}^{0}1\times x+{\int}_{-3}^{0}\sqrt{9-{x}^{2}}dx$

$I={A}_{1}+{A}_{2}$

we can plot area A1 as follows

we can plot area A2 as follows

Hence we can calculate the integral as

$I=W\times H+\frac{\pi {r}^{2}}{4}$

$I=3\times 1+\frac{\pi {\left(3\right)}^{2}}{4}$

$I=3+\frac{9\pi}{4}$

The given integral is

we can plot area A1 as follows

we can plot area A2 as follows

Hence we can calculate the integral as

nick1337

Expert

2021-12-28Added 573 answers

The expression you are having trouble with seems to be incorrect.

You have

Since

you have

Applying your definition:

From this, we see that

Then the final expression should be

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