Evaluate the improper integral.
-$$\int _ { 0 } ^ { 4 } x \ln 4 x d x$$
A) $8 \ln 16 - 4$
B) $- \frac { 1 } { 2 } \ln 16 + 4$
C) $8 \ln 4 - 4$
D) $8 \ln 16 - 8$

Question

Quizplus · Accepted Answer

To solve the integral $\int _ { 0 } ^ { 4 } x \ln 4 x d x$, we can use integration by parts, where we let $u = \ln 4x$ and $dv = x dx$. This gives us $du = \frac{1}{x} dx$ and $v = \frac{1}{2}x^2$. Applying the integration by parts formula, $\int u dv = uv - \int v du$, we get:$$\frac{1}{2}x^2 \ln 4x - \int \frac{1}{2}x^2 \cdot \frac{1}{x} dx = \frac{1}{2}x^2 \ln 4x - \frac{1}{4}x^2$$Evaluating this from 0 to 4, we get:$$\left(\frac{1}{2} \cdot 4^2 \ln 16 - \frac{1}{4} \cdot 4^2\right) - \left(0 - 0\right) = 8 \ln 16 - 4$$Therefore, the correct answer is $8 \ln 16 - 4$.

Evaluate the Improper Integral ∫04xln⁡4xdx\int _ { 0 } ^ { 4 } x \ln 4 x d x∫04​xln4xdx

Evaluate the Improper Integral $\int _ { 0 } ^ { 4 } x \ln 4 x d x$