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NDA 2023 Maths · Mathematics · Definite Integrals · medium

Let f(x)=x for 0≤ x≤1 and f(x)=2-x for 1<x≤2. Then ∫0^2f(x) dx is equal to

Single correct option

Let f(x)=xf(x)=x for 0≤x≤10\le x\le1 and f(x)=2−xf(x)=2-x for 1<x≤21<x\le2. Then ∫02f(x) dx\displaystyle\int_0^2f(x)\,dx is equal to
  1. A. 22
  2. B. 12\dfrac12
  3. C. 32\dfrac32
  4. D. 11
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Answer: (D) 11

The graph is a triangle with base 22 and height 11, so the integral is 12⋅2⋅1=1\tfrac12\cdot2\cdot1=1. Algebraically, ∫01x dx+∫12(2−x)dx=12+12=1\int_0^1x\,dx+\int_1^2(2-x)dx=\dfrac12+\dfrac12=1. The value 12\dfrac12 takes only one piece.
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