Question:
The value of the integral \(∫^2_{-2}\frac{ |x^3+x|}{(e^x|x|+1)}dx\) is equal to:
The value of the integral \(∫^2_{-2}\frac{ |x^3+x|}{(e^x|x|+1)}dx\) is equal to:
Updated On: Oct 6, 2024
- \(5e^2\)
- \(3e^{–2}\)
- 4
- 6
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The Correct Option is D
Solution and Explanation
\(I = ∫^2_{-2}\frac{ |x^3+x|}{(e^x|x|+1)}dx.....(i)\)
\(I = ∫^2_{-2}\frac{ |x^3+x|}{(e^x|x|+1)}dx.....(ii)\)
\(2I = ∫^2_{-2} |x^3+x|dx\)
\(2I = 2 ∫^2_0(x^3+x)dx\)
\(I = ∫^2_0 (x^3+x)dx\)
= \(\bigg(\frac{16}{4}+\frac{4}{2}\bigg)-0\)
= \(4+2 = 6\)
Hence, the correct option is (D): \(6\)
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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.