Question:
If $f(x) = x^2 , g(x) = 2x,0 \leq x \leq 2$ then the value of $I(x) = \int\limits_0^2 max (f(x), g(x))$ is
If $f(x) = x^2 , g(x) = 2x,0 \leq x \leq 2$ then the value of $I(x) = \int\limits_0^2 max (f(x), g(x))$ is
Updated On: Apr 18, 2024
- $\frac{10}{3}$
- $\frac{1}{3}$
- $\frac{11}{3}$
- 32
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The Correct Option is D
Solution and Explanation
Let $r\left(x\right) = f\left(x\right).g\left(x\right)$
$ =x^{2} .2x = 2x^{3}$
$ r'\left(x\right) = 6x^{2} $
Put $6x^{2} = 0 , \, \, \therefore x = 0$
Max $ r\left(x\right)=2\left(2\right)^{3} = 16$
or Max $ \left(f\left(x\right),g\left(x\right)\right) = 16 $
$ I\left(x\right)= \int^{2}_{0} 16dx$
$ I\left(x\right) = \left[16x\right]^{2}_{0} = 32 -0=32$
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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.