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JEE Main 2020 · Mathematics · Indefinite and Definite Integrals · medium

Find the value of 3∫0^3|x^2-2x| dx.

Numerical answer

Find the value of 3∫03∣x2−2x∣ dx3\displaystyle\int_0^3|x^2-2x|\,dx.
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Answer: 8

x2−2x<0x^2-2x<0 on (0,2)(0,2) and >0>0 on (2,3)(2,3). ∫02(2x−x2)dx=4−83=43\int_0^2(2x-x^2)dx=4-\tfrac83=\tfrac43 and ∫23(x2−2x)dx=(9−9)−(83−4)=43\int_2^3(x^2-2x)dx=(9-9)-(\tfrac83-4)=\tfrac43. Total =83=\tfrac83, so 3×83=83\times\tfrac83=8.
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