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Solve $\int_{-2}^{2}|1-x^{2}\lvert.\text{d}x$ [IIT 1989; BIT Mesra 1996; Kurukshetra CEE 1998; MP PET 2002; Kerala (Engg.) 2002]
(A) 2
(B) 4
(C) 6
(D) 8

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Hint: Integration means adding smaller functions to create a larger one. It is the inverse of differentiation, so it is also known as anti differentiation. Integrals with no integration limit are known as indefinite integrals. Integrals with an upper and lower limit are said to be definite integrals.

Complete step by step solution:The given integral is I = $\int_{-2}^{2}|1-x^{2}\lvert.\text{d}x$
Here the given function is f(x) = $\|1-x^{2}\lvert$ and the total range over which the integral is to be done is -2 to 2, with a lower limit of -2 and an upper limit of 2. We observe that the value of this function is negative in the range of -2 to -1, positive in the range of -1 to 1, and again negative in the range of 1 to 2.
So, the above integral can be written as
I = $-\int_{-2}^{-1}(1-x^{2}).\text{d}x+\int_{-1}^{1}(1-x^{2}).\text{d}x-\int_{1}^{2}(1-x^{2}).\text{d}x$
I = $-\left[x-\dfrac{x^3}{3}\right]_{-2}^{-1}+\left[x-\dfrac{x^3}{3}\right]_{-1}^{1}-\left[x-\dfrac{x^3}{3}\right]_{1}^{2}$
I = $-((-1)-\dfrac{-1^3}{3}-((-2)-\dfrac{-2^3}{3}))+((1)-\dfrac{1^3}{3}-((-1)-\dfrac{-1^3}{3})+((2)-\dfrac{2^3}{3}-((1)-\dfrac{1^3}{3}))$
I = $-(-1+\dfrac{1}{3}+2-\dfrac{8}{3})+(1-\dfrac{1}{3}+1-\dfrac{1}{3})-(2-\dfrac{8}{3}-1+\dfrac{1}{3})$
I = $\dfrac{4}{3}+\dfrac{4}{3}+\dfrac{4}{3}$
I = 4
Hence, the integration of $\|1-x^{2}\lvert$over the range of -2 to 2 is 4.

Option ‘B’ is correct

Note: The integration should be done carefully. Integration can be applied in daily life. It is used in chemistry to study radioactive decay reactions. It can be used to calculate the velocity and trajectory of the object. It is also used to calculate the centre of mass, centre of gravity, mass, and momentum of the satellites.

Last updated date: 30th Sep 2023
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