Question:
The value of the integral, $\int_\limits{1}^{3}\left[ x ^{2}-2 x -2\right] dx ,$
where $[x]$ denotes the greatest integer less than or equal to $x$, is :
The value of the integral, $\int_\limits{1}^{3}\left[ x ^{2}-2 x -2\right] dx ,$
where $[x]$ denotes the greatest integer less than or equal to $x$, is :
Updated On: Feb 14, 2025
- $-\sqrt{2}-\sqrt{3}+1$
- $-\sqrt{2}-\sqrt{3}-1$
- -5
- -4
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The Correct Option is B
Solution and Explanation
$\int_\limits{1}^{3}\left(\left[(x-1)^{2}\right]-3\right) d x $
$=\int_\limits{1}^{2}\left[x^{2}\right]-3 \int_\limits{1}^{3} d x $
$=\int_\limits{1}^{3} 0 d x+\int_\limits{1}^{\sqrt{2}} 1 .d x+\int_\limits{\sqrt{2}}^{\sqrt{3}} 2.d x+\int_\limits{\sqrt{3}}^{2} 3. d x-6$
$=\sqrt{2}-1+2(\sqrt{3}-\sqrt{2})+3(2-\sqrt{3})-6 $
$=-\sqrt{2}-\sqrt{3}-1$
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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.