Answer
Proof
Work Step by Step
To show that :- \[\int_{0}^{1}x^2\cos x\:dx\leq \frac{1}{3}\]
Let $f(x)=x^2\cos x$ on $[0,1]$
\[|f(x)|=|x^2\cos x|=|x^2||\cos x|\]
We know that \[|\cos x|\leq 1\]
\[\Rightarrow |f(x)|=|x^2||\cos x|\leq x^2\]
\[\Rightarrow |f(x)|\leq x^2\] on $[0,1]$
We will use the result , If $F(x)\leq G(x)$ on $[a,b]$ then
\[\int_{a}^{b}F(x)\:dx\leq \int_{a}^{b}G(x)\:dx\;\;\;...(1)\]
By using (1)
\[\Rightarrow \int_{0}^{1}|f(x)|\:dx\leq \int_{0}^{1}x^2\:dx\]
\[\Rightarrow \int_{0}^{1}|f(x)|\:dx\leq \int_{0}^{1}x^2\:dx=\left[\frac{x^3}{3}\right]_{0}^{1}\]
\[\Rightarrow \int_{0}^{1}|f(x)|\:dx\leq \frac{1}{3}\]
Since $f(x)\leq |f(x)|$ on $[0,1]$
Using (1) with $F(x)=f(x)$ and $G(x)=|f(x)|$
\[\Rightarrow \int_{0}^{1}f(x)\:dx\leq \int_{0}^{1}|f(x)|\:dx\leq \frac{1}{3}\]
\[\Rightarrow \int_{0}^{1}f(x)\:dx\leq\frac{1}{3}\]
Hence proven.
