Question:
$ \int\limits_{0} ^{1}\frac{dx}{[ax+b(1-x)]^2}$ is equal to
$ \int\limits_{0} ^{1}\frac{dx}{[ax+b(1-x)]^2}$ is equal to
Updated On: Aug 10, 2024
- ab
- $\frac{a}{b}$
- $\frac{b}{a}$
- $\frac{1}{ab}$
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The Correct Option is B
Solution and Explanation
$\int\limits_{0}^{1} \frac{dx}{\left[ax+b\left(1-x\right)\right]^{2}}= \int\limits_{0}^{^1}\left[b+\left(a-b\right)x\right]^{-2} dx$
$=\left|\frac{b+\left(a-b\right)x^{-1}}{-\left(a-b\right)}\right|_{0}^{1} $
$=-\frac{1}{a-b} \left([b+\left(a-b\right)\right]^{-1}-\left[b\right]^{-1}])$
$=-\frac{1}{a-b}\left[\frac{1}{a}-\frac{1}{b}\right]$
$=-\frac{1}{a-b}\cdot \frac{b-a}{ab}=\frac{1}{ab}$
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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.