Question:
$ \int\limits_{2-a}^{2+a} f\left(x\right) dx $ is equal to (where $f\left(2 - a\right)=f\left(2 + a\right) \forall a\in R$)
$ \int\limits_{2-a}^{2+a} f\left(x\right) dx $ is equal to (where $f\left(2 - a\right)=f\left(2 + a\right) \forall a\in R$)
Updated On: Jul 6, 2022
- $ 2 \int\limits_{2}^{2+a} f\left(x\right) dx$
- $ 2 \int\limits_{0}^{a} f\left(x\right) dx$
- $ 2 \int\limits_{0}^{2} f\left(x\right) dx$
- None of these
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The Correct Option is A
Solution and Explanation
$ \because f\left(2 - a \right)=f\left(2 + a\right) $
$\therefore$ Function is symmetrical about the line $x = 2$
Then $ \int\limits_{2-a}^{2+a} f\left(x\right) dx = 2 \int\limits_{2}^{2+a} f\left(x\right) dx$
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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.