Question

${}_0^5\int \frac{4\sqrt{x+4}}{4\sqrt{x+4}+4\sqrt{9-x}}$

Answer 1

Consider the given integral.

$I={\int }_0^5\frac{4\sqrt{x+4}}{4\sqrt{x+4}+4\sqrt{9-x}}dx$ ……. (1)

We know that

${\int }_a^{b}f\left(x\right)dx={\int }_a^{b}f(a+b-x)dx$

Therefore,

$I={\int }_0^5\frac{4\sqrt{5-x+4}}{4\sqrt{5-x+4}+4\sqrt{9-(5-x)}}dx$

$I={\int }_0^5\frac{4\sqrt{9-x}}{4\sqrt{9-x}+4\sqrt{4+x}}dx$

$I={\int }_0^5\frac{4\sqrt{9-x}}{4\sqrt{9-x}+4\sqrt{x+4}}dx$ …… (2)

On adding equation (1) and (2), we get

$I={\int }_0^5\frac{4\sqrt{9-x}+4\sqrt{x+4}}{4\sqrt{9-x}+4\sqrt{x+4}}dx$

$I={\int }_0^{5}1dx$

$I={\left[x\right]}_0^5$

$I=5-0$

$I=5$

Hence, this is the answer.