Question

How do you evaluate the definite integral $\int {(\sqrt {4 - {x^2}} } )dx$ from $[ 2,2]?$

Answer 1

Hint:As we know that in integration there are two types of integrals. They are definite integrals and indefinite integrals. Definite integrals are those integrals which have an upper and lower limit and have different values for upper and lower limit when they are evaluated.

 The final value of definite integral is the value of integral to the upper limit minus value of the definite integral for the lower limit which is as, $\int\limits_a^b {f(x)*dx = F(b) - F(a)} $, here $F(b)$and $F(a)$are the integral function for the upper and lower limit.

Complete step by step solution:
Here we will use the substitution method to solve the given question, $x = 2\sin \theta ,dx = 2\cos \theta d\theta $

Now,$\sqrt {4 - {x^2}} = \sqrt {4 - 4{{\sin }^2}\theta } = 2\cos \theta $,$\int {\sqrt {4 - {x^2}} } dx = \int {2\cos \theta *2\cos \theta d\theta \Rightarrow 4\int {{{\cos }^2}} } \theta d\theta $

We know that $\dfrac{{\cos 2\theta + 1}}{2} = {\cos ^2}\theta $cos 2, We have:$4\int {{{\cos }^2}}

\theta d\theta = 2\int {(\cos 2\theta + 1)d\theta = 2\dfrac{{\sin 2\theta }}{2}} + 2\theta

\Rightarrow 2\sin \theta \cos \theta + 2\theta $

It gives us $2*\dfrac{x}{2}*\dfrac{{\sqrt {4 - {x^2}} }}{2} + 2\arcsin (\dfrac{x}{2}) = x*\dfrac{{\sqrt
{4 - {x^2}} }}{2} + 2\arcsin (\dfrac{x}{2}) + C$

Therefore, $\int\limits_{ - 2}^2 {\sqrt {4 - {x^2}} dx = [x - \dfrac{{\sqrt {4 - {x^2}} }}{2} + 2\arcsin (\dfrac{x}{2}} )]$$ \Rightarrow 2*\dfrac{\pi }{2} + 2*\dfrac{\pi }{2} = 2\pi $.

Hence the required answer is $2\pi $.

Note: While calculating definite integral we should keep in mind about the upper and lower limit. Lower limit of the first addend should be equal to the lower limit of the original definite integral. It is an arbitrary constant.

 In definite integral the function $f(x)$is considered as the integrand and $dx$ is the integrating agent. We should know that a definite integral is also called Riemann integral. Indefinite integrals are those integrals which do not have any limit of integration as in the above the question we have been provided with upper and lower limits so it was a definite integral.

There are several properties of definite integral one of them is that a definite integral can also be written as sum of two definite integrals.