Question:
$ \int \limits_1^3|(2-x) \log x|dx $ is equal to
$ \int \limits_1^3|(2-x) \log x|dx $ is equal to
Updated On: Jul 6, 2022
- $\frac{3}{2} \log \, 3 + \frac{1}{2}$
- $\log \frac{16}{3 \sqrt{3}}- \frac{1}{2}$
- $-\frac{3}{2} \log 3 - \frac{1}{2}$
- none of these.
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The Correct Option is B
Solution and Explanation
$\int\limits_{1}^{3} \left|\left(2-x\right)log\,x\right|dx=\int\limits_{1}^{2}\left|2-x\right|\left|log\,x\right|dx$
$+\int\limits_{2}^{3}\left|2-x\right|\left|log\,x\right|dx$
$=\int\limits_{1}^{2}\left(2-x\right)log \, x\,dx+\int\limits_{2}^{3}\left(x-2\right)log\,x \, dx$
$=\left|log\,x\left(2x-\frac{x^{2}}{2}\right)\right|_{1}^{2}-\int_{1}^{2} \frac{1}{x} \left(2x-\frac{x^{2}}{2}\right)dx$
$+\left|log\,x\left(\frac{x^{2}}{2}-2x\right)\right|_{2}^{3}-\int_{2}^{3} \frac{1}{x}\left(x^{2}-2x\right)dx$
$=4 log\, 2-\frac{3}{2} log 3 -\frac{1}{2}$
$=log \frac{16}{3\sqrt{3}}-\frac{1}{2}$
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Concepts Used:
Integral
The representation of the area of a region under a curve is called to be as integral. The actual value of an integral can be acquired (approximately) by drawing rectangles.
- The definite integral of a function can be shown as the area of the region bounded by its graph of the given function between two points in the line.
- The area of a region is found by splitting it into thin vertical rectangles and applying the lower and the upper limits, the area of the region is summarized.
- An integral of a function over an interval on which the integral is described.
Also, F(x) is known to be a Newton-Leibnitz integral or antiderivative or primitive of a function f(x) on an interval I.
F'(x) = f(x)
For every value of x = I.
Types of Integrals:
Integral calculus helps to resolve two major types of problems:
- The problem of getting a function if its derivative is given.
- The problem of getting the area bounded by the graph of a function under given situations.