Question

Find the area of the circle \[{x^2} + {y^2} = {a^2}\]by integration

Answer 1

Hint: In mathematics an integral assign no. To functions in a way that can describe displacement area, volume and other concepts that arise by combining infinitesimal data. Integration is one of the two main operations of calculus its inverse operation is differentiation
\[\int_a^b {f(x)dx = \left[ {f(x)} \right]_a^b} = f(b) - f(a)\]
Integration is way of adding slices to fin the whole integration can be used to find areas volumes central points and many useful things

Complete step by step answer:
\[{x^2} + {y^2} = {a^2}\] dimension of a circle
centre\[ = (0,0)\]
Radius\[ = a\]
Hence \[OA = OB = \] radius\[ = a\] (assume)
\[A = \left[ {a,o} \right]\]
\[B = \left[ {o,a} \right]\]
Area of circle \[ = 4x\] area of region OBA
\[4x\int_b^a {ydx} \]
We know that
\[{x^2} + {y^2} = {a^2}\]
\[{y^2} = {a^2} - {x^2}\]
\[y = \sqrt[ + ]{{{a^2} - {x^2}}}\]
Sine AOBA lies in Ist quadrant value of y us fibe
\[y = \sqrt[{}]{{{a^2} - {x^2}}}\]
Now
Area of circle \[4x\int_b^a {\sqrt {{a^2} - {x^2}dx} } \]
Using integration
\[\sqrt {{a^2} - {x^2}} dx = \dfrac{1}{2}{x^2} - {x^2} + \dfrac{{{a^2}}}{2}{\sin ^{ - 1}}\dfrac{x}{a} + c\]
\[ = 4\left[ {\dfrac{x}{2}\sqrt {{a^2} - {x^2}} + \dfrac{{{a^2}}}{2}{{\sin }^{ - 1}}\dfrac{x}{a}} \right]_o^a\]
\[ = z{a^2}\]
\[ = 4\left[ {o + \dfrac{{{a^2}}}{2}{{\sin }^{ - 1}}(1) - o - o} \right]\]
\[ = 4 \times \dfrac{{{a^2}}}{2}{\sin ^{ - 1}}(1)\]
\[ = 4 \times \dfrac{{{a^2}}}{2} \times \dfrac{z}{2}\]
\[ = z{a^2}\]

Note:
 In calculus integration by substitution also known as u-substitution or change of variables is a method of evaluating integrals. Direct application of the fundamental theorem of calculus to find an ant derivative can be quite difficult and integration by substitution can help simplify that task.