Question

The function $f(x)=\int_{e^{x}}^{e^{2x}} t \log_{e}t \, dt$ has an absolute minima at $x=0$ and a local maxima at $x=$

• A. $-\log_{e}4$
• B. $\log_{e}2$
• C. 1
• D. $-\log_{e}2$

Hint:

Leibniz rule is essential whenever the limits of an integral are functions of the variable you are differentiating with respect to.

Answer 1

The Correct Option is D

This problem requires the Leibniz Rule for differentiating under an integral sign to find critical points.
Step 1: The derivative of $f(x) = \int_{g(x)}^{h(x)} w(t) dt$ is $f'(x) = w(h(x))h'(x) - w(g(x))g'(x)$. Here $w(t) = t \log t$. $f'(x) = [e^{2x} \log(e^{2x})] \cdot (2e^{2x}) - [e^x \log(e^x)] \cdot (e^x)$. $f'(x) = [e^{2x} \cdot 2x] \cdot 2e^{2x} - [e^x \cdot x] \cdot e^x$. $f'(x) = 4x e^{4x} - x e^{2x}$.
Step 2: Set the first derivative to zero: $4x e^{4x} - x e^{2x} = 0$. Factor out common terms: $x e^{2x} (4 e^{2x} - 1) = 0$. This gives two possible equations: 1) $x = 0$ (Given as the absolute minimum). 2) $4 e^{2x} - 1 = 0$.
Step 3: $4 e^{2x} = 1 \implies e^{2x} = \frac{1}{4}$. Take the natural logarithm of both sides: $2x = \log_e(1/4)$. Using log properties: $2x = \log_e(4^{-1}) = -\log_e(4) = -2\log_e(2)$. Dividing by 2 gives: $x = -\log_e 2$.
Step 4: At $x = -\log_e 2$, we check the sign of $f'(x)$ around this value. For $x < -\log_e 2$, $f'(x)$ is positive. For $x > -\log_e 2$ (but less than 0), $f'(x)$ is negative. Since the derivative changes from positive to negative, $x = -\log_e 2$ is a local maximum. The correct option is (4).