## Calculus: Early Transcendentals 8th Edition

$$\int_{0}^{1} x \sin^{-1}x d x$$ Since $$0 \leq x \leq 1 \Rightarrow 0 \leq \sin^{-1}x \leq \frac{\pi}{2} \Rightarrow x\sin^{-1}x \leq x (\frac{\pi}{2})$$ then we have : $$\int_{0}^{1} x \sin^{-1}x d x \leq \int_{0}^{1} x (\frac{\pi}{2}) d x= \left[\frac{\pi}{4}x^{2}\right]_{0}^{1}=\frac{\pi}{4} [\text { Property } 7]$$
$$\int_{0}^{1} x \sin^{-1}x d x$$ Since $$0 \leq x \leq 1 \Rightarrow 0 \leq \sin^{-1}x \leq \frac{\pi}{2} \Rightarrow x\sin^{-1}x \leq x (\frac{\pi}{2})$$ then we have : $$\int_{0}^{1} x \sin^{-1}x d x \leq \int_{0}^{1} x (\frac{\pi}{2}) d x= \left[\frac{\pi}{4}x^{2}\right]_{0}^{1}=\frac{\pi}{4} [\text { Property } 7]$$