Find the area of the region bounded by x2=4y,y=2,y=4 and the x-axis in the first quadrant.

The area of the region bounded by the curve, x2 = 4y,y=2,and y=4,and the y-axis
is the area ABCD.
Area of ABCD=\(\int^4_2xdy\)
=\(\int^4_2 2\sqrt ydy\)
=\(2\int^4_2 \sqrt ydy\)
=\(2\bigg[\frac{y^{\frac{3}{2}}}{\frac{3}{2}}\bigg]^4_2\)
=\(\frac{4}{3}\bigg[(4)^{\frac{3}{2}}-(2)^{\frac{3}{2}}\bigg]\)
=\(\frac{4}{3}\)[8-2\(\sqrt 2\)]
=\(\bigg(\frac{32-8\sqrt 2}{3}\bigg)\)units
Let \( f : (0, \infty) \to \mathbb{R} \) be a twice differentiable function. If for some \( a \neq 0 \), } \[ \int_0^a f(x) \, dx = f(a), \quad f(1) = 1, \quad f(16) = \frac{1}{8}, \quad \text{then } 16 - f^{-1}\left( \frac{1}{16} \right) \text{ is equal to:}\]
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.
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.
Integral calculus helps to resolve two major types of problems: