Let $\displaystyle f'(x)=\frac{192x^3}{2+\sin^4(\pi x)}\forall x\in\mathbb{R}$ with $f(0.5)=0.$ If $$ m\leq \int^{1}_{0.5}f(x)dx\leq M,$$ then possible values of $m,M$ are
$(a)\; 13,24\;\;\;\;\;\; (b)\;0.25,0.5\;\;\;\; (c)\;\; -11,0\;\;\;\; (d)1,12$
Try: $\displaystyle f'(x)=\frac{192x^3}{2+\sin^4(\pi x)}>0$ in $(0.5,1)$ so $f(x)$ is a strictly increasing function. Could someone help me to solve it. Thanks!
For $x$ in $[0.5,1]$, $$ f(x)=f(0.5)+\int_{0.5}^x f'(t)dt=\int_{0.5}^x f'(t)dt. $$ Moreover, since $\sin$ varies between $-1$ and $1$, we know that its fourth power varies between $0$ and $1$, so $$ \frac{192x^3}{3}\leq f'(x)\leq\frac{192x^3}{2}. $$ Therefore, $$ \int_{0.5}^x 64t^3dt\leq f(x)\leq \int_{0.5}^x 96t^3dt. $$ and so $$ 16(x^4-0.5^4)\leq f(x)\leq 24(x^4-0.5^4). $$ Substituting this into the original expression, we have that $$ \int_{0.5}^1 16(x^4-0.5^4)dx\leq \int_{0.5}^1 f(x)dx\leq \int_{0.5}^1 24(x^4-0.5^4)dx $$ Evaluating the integrals gives $$ 2.6\leq \int_{0.5}^1 f(x)dx\leq 3.9 $$ Therefore, I get that $(d)$ is the right answer.