Find the area of the region bounded by y2=9x, x=2, x=4 and the x-axis in the first quadrant.
Find the area of the region bounded by y2=9x, x=2, x=4 and the x-axis in the first quadrant.
Solution and Explanation
The area of the region bounded by the curve,y2=9x,x=2,and x=4,and the x-axis is the area ABCD.
Area of ABCD =\(\int^4_2ydx\)
=\(\int^4_2\sqrt xdx\)
=\(3\bigg[\frac{x^{\frac{3}{2}}}{\frac{3}{2}}\bigg]\)
=\(2\bigg[x^{\frac{3}{2}}\bigg]^4_2\)
=\(2\bigg[(4)^{\frac{3}{2}}-(2)^{\frac{3}{2}}\bigg]\)
=\(2[8-2\sqrt2]\)
=(16-4\(\sqrt2\)) units.
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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.