JEE MAIN - Mathematics (2020 - 3rd September Morning Slot - No. 11)
$$\int\limits_{ - \pi }^\pi {\left| {\pi - \left| x \right|} \right|dx} $$ is equal to :
$${\pi ^2}$$
2$${\pi ^2}$$
$$\sqrt 2 {\pi ^2}$$
$${{{\pi ^2}} \over 2}$$
Explanation
$$\int\limits_{ - \pi }^\pi {\left| {\pi - \left| x \right|} \right|dx} $$
= $$2\int\limits_0^\pi {\left| {\pi - \left| x \right|} \right|} dx$$ [As it is even function]
= $$2\int\limits_0^\pi {\left( {\pi - x} \right)} dx$$
= $$2\left[ {\pi x - {{{x^2}} \over 2}} \right]_0^\pi $$
= $${\pi ^2}$$
= $$2\int\limits_0^\pi {\left| {\pi - \left| x \right|} \right|} dx$$ [As it is even function]
= $$2\int\limits_0^\pi {\left( {\pi - x} \right)} dx$$
= $$2\left[ {\pi x - {{{x^2}} \over 2}} \right]_0^\pi $$
= $${\pi ^2}$$
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