# Determine the following definite integrals: \int_{-5}^{0}(1+\sqrt{25-x^{2}})dx

Determine the following definite integrals:
$$\displaystyle{\int_{{-{5}}}^{{{0}}}}{\left({1}+\sqrt{{{25}-{x}^{{{2}}}}}\right)}{\left.{d}{x}\right.}$$

• Questions are typically answered in as fast as 30 minutes

### Plainmath recommends

• Get a detailed answer even on the hardest topics.
• Ask an expert for a step-by-step guidance to learn to do it yourself.

Aniqa O'Neill
Evaluate the integrals:
$$\displaystyle{f{{\left({x}\right)}}}={\int_{{-{5}}}^{{{0}}}}{\left({1}+\sqrt{{{25}-{x}^{{{2}}}}}\right)}{\left.{d}{x}\right.}$$
In terms of areas:
$$\displaystyle{f{{\left({x}\right)}}}={\int_{{-{5}}}^{{{0}}}}\sqrt{{{25}-{x}^{{{2}}}}}{\left.{d}{x}\right.}$$
Let $$\displaystyle{x}={5}{\sin{{u}}}$$
$$\displaystyle{f{{\left({x}\right)}}}={{\left[{x}\right]}_{{-{5}}}^{{{0}}}}+{\int_{{-{\frac{{\pi}}{{{2}}}}}}^{{{0}}}}\sqrt{{{25}-{\left({5}{\sin{{u}}}\right)}^{{{2}}}}}{d}{u}$$
$$\displaystyle={{\left[{x}\right]}_{{-{5}}}^{{{0}}}}+{\int_{{-{\frac{{\pi}}{{{2}}}}}}^{{{0}}}}{25}{{\cos}^{{{2}}}{\left({u}\right)}}{d}{u}$$
$$\displaystyle={\left[{0}-{\left(-{5}\right)}\right]}+{25}\cdot{\int_{{-{\frac{{\pi}}{{{2}}}}}}^{{{0}}}}{{\cos}^{{{2}}}{\left({u}\right)}}{d}{u}$$
$$\displaystyle={5}+{25}\cdot{\frac{{{1}}}{{{2}}}}\cdot{\int_{{{\frac{{\pi}}{{{2}}}}}}^{{{0}}}}{1}+{\cos{{\left({2}{u}\right)}}}{d}{u}$$
$$\displaystyle={5}+{\frac{{{25}}}{{{2}}}}{\left[{{\left[{u}\right]}_{{-{\frac{{\pi}}{{{2}}}}}}^{{{0}}}}+{{\left[{\frac{{{1}}}{{{2}}}}{\sin{{2}}}{u}\right]}_{{-{\frac{{\pi}}{{{2}}}}}}^{{{0}}}}\right]}$$
$$\displaystyle={5}+{\frac{{{25}}}{{{2}}}}{\left[{\left[{0}-{\left(-{\frac{{\pi}}{{{2}}}}\right)}\right]}+{\left({0}\right)}\right]}$$
$$\displaystyle={5}+{\frac{{{25}}}{{{2}}}}{\left[{\frac{{\pi}}{{{2}}}}+{0}\right]}$$
$$\displaystyle={5}+{\frac{{{25}\pi}}{{{4}}}}$$