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The value of improper integral

$\displaystyle\int_{0}^{1} x\ln x =?$

1. $1/4$
2. $0$
3. $-1/4$
4. $1$

Using the $ILATE$ rule to select first and second function for integrating it by parts: The first function should be Logarithmic and second should be Algebraic.

For this question, first function would be  $ln(x)$ and second would be $x$

$\therefore$ $\int_{0}^{1}xln(x)dx=ln(x)\times \frac{x^{2}}{2}-\int \frac{1}{x}\times \frac{x^{2}}{2}dx$

$\Rightarrow$  $\int_{0}^{1}xln(x)dx=ln(x)\times \frac{x^{2}}{2}-\frac{x^{2}}{4}$

$\Rightarrow$  $\int_{0}^{1}xln(x)dx=\left [ {0-\frac{1}{4}}-0 \right ]$ $=\frac{-1}{4}$

Option C is correct.
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### 1 comment

Although the answer is right but the way you handled improper integral is wrong

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