IIT-JEE 1997Question 1
Let $a+b=4$, where $a<2$, and let $g(x)$ be a differentiable function. If $\dfrac{dg}{dx}>0$ for all $x$, prove that $\displaystyle\int_0^{a}g(x)\,dx+\int_0^{b}g(x)\,dx$ increases as $(b-a)$ increases.
Definite Integration
Answerproved
Solution
Concept. Write the sum as a function of $a$ alone ($b=4-a$) and look at the sign of its derivative. Formulas used. (1) $\frac{d}{da}\int_0^{a}g=g(a)$, $\ \frac{d}{da}\int_0^{4-a}g=-g(4-a)$(2) $g'>0\Rightarrow g$ is strictly increasingStep 1 - Set up. $\phi(a)=\int_0^ag(x)\,dx+\int_0^{4-a}g(x)\,dx$, and $b-a=4-2a$. Step 2 - Derivative. By (1), $\phi'(a)=g(a)-g(4-a)$. Step 3 - Sign. $a<2$ gives $a<4-a$, so by (2) $g(a)<g(4-a)$ and $\phi'(a)<0$. So $\phi$ decreases as $a$ increases. Step 4 - Conclude. $b-a=4-2a$ increases exactly when $a$ decreases, and then $\phi$ increases. Hence proved. Answer: proved