Question:
$ \int\limits_{5}^{10} \frac{1}{\left(x-1\right)\left(x-2\right)}dx $ is equal to
$ \int\limits_{5}^{10} \frac{1}{\left(x-1\right)\left(x-2\right)}dx $ is equal to
Updated On: Aug 17, 2024
- $ log \frac{27}{32} $
- $ log \frac{32}{27} $
- $ log \frac{8}{9} $
- $ log \frac{3}{4} $
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The Correct Option is B
Solution and Explanation
Let $ I =\int \limits_{5}^{10} \frac{1}{(x-1)(x-2)} d x $
$=\int \limits_{5}^{10}\left[\frac{-1}{x-1}+\frac{1}{x-2}\right] d x $
$=[-\log (x-1)+\log (x-2)]_{5}^{10} $
$=-\log 9+\log 8+\log 4-\log 3 $
$=-2 \log 3+3 \log 2+2 \log 2-\log 3 $
$=-3 \log 3+5 \log 2 $
$=-\log 27+\log 32 $
$=\log \frac{32}{27} $
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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.