## Calculus: Early Transcendentals 8th Edition

$$\int^{1}_0(\frac{4}{1+p^2})dp=\pi$$
$$A=\int^{1}_0(\frac{4}{1+p^2})dp$$ $$A=4\int^{1}_0(\frac{1}{1+p^2})dp$$ According to Table 1, we have $$\int(\frac{1}{x^2+1})dx=\tan^{-1}x+C$$ Therefore, $$A=4(\tan^{-1}p)\Bigg]^1_0$$ $$A=4(\tan^{-1}1-\tan^{-1}0)$$ $$A=4(\frac{\pi}{4}-0)$$ (we know that $\tan x=\frac{\sin x}{\cos x}$. So for $\tan x$ to equal $0$, $\sin x$ must also equal $0$, and we already know $\sin 0=0$. Therefore, $\tan^{-1}0=0$) $$A=4\times\frac{\pi}{4}$$ $$A=\pi$$