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Question Number 201762 by Calculusboy last updated on 11/Dec/23

Answered by mr W last updated on 12/Dec/23

$${\int }_{-2}^2(x^3\cos \frac{x}{2}+\frac{1}{2})\sqrt{4-x^2}dx$$$$={\int }_{-2}^2\underset{oddfunction}{\left(x^3\cos \frac{x}{2}\right)\sqrt{4-x^2}}dx+{\int }_{-2}^2\underset{evenfunction}{\frac{1}{2}\sqrt{4-x^2}}dx$$$$=0+{\int }_0^2\sqrt{4-x^2}dx$$$$=\frac{\pi \times 2^2}{4}=\pi ✓$$

Commented by Calculusboy last updated on 11/Dec/23

$$\mathit{t}\mathit{h}\mathit{a}\mathit{n}\mathit{k}\mathit{s}\mathit{s}\mathit{i}\mathit{r}$$

Commented by mr W last updated on 12/Dec/23

$$iff(-x)=-f\left(x\right),then$$$${\int }_{-a}^{a}f\left(x\right)dx=0.$$$$iff(-x)=f\left(x\right),then$$$${\int }_{-a}^{a}f\left(x\right)dx=2{\int }_0^{a}f\left(x\right)dx$$

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