## University Calculus: Early Transcendentals (3rd Edition)

$$\int^1_0 x\sqrt{1-x}dx=\frac{4}{15}$$
$$A=\int^1_0 x\sqrt{1-x}dx$$ Let $n=\sqrt{1-x}$, which means $n^2=1-x$ and $x=1-n^2$. Also, $dn=\frac{(1-x)'}{2\sqrt{1-x}}dx=\frac{-1}{2\sqrt{1-x}}dx=\frac{-1}{2n}dx$. Therefore, $dx=-2ndn$ For $x=0$, we have $n=\sqrt{1-0}=1$ and for $x=1$, we have $n=\sqrt{1-1}=0$ $$A=\int^0_1 (1-n^2)n(-2ndn)$$ $$A=\int^0_1(1-n^2)(-2n^2)dn$$ $$A=\int^0_1(2n^4-2n^2)dn$$ $$A=\Big(\frac{2n^5}{5}-\frac{2n^3}{3}\Big)\Bigg]^0_1$$ $$A=0-\Big(\frac{2}{5}-\frac{2}{3}\Big)$$ $$A=-\Big(-\frac{4}{15}\Big)=\frac{4}{15}$$