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NDA 2024
Exam
NDA
Year
2024
Subject
Mathematics
Chapter
Definite Integration

NDA 2024 · Mathematics · Definite Integration

Let f(x)=|x-1|, g(x)=[x] and h(x)=f(x) g(x) where [.] is greatest...

Let f(x)=∣x−1∣,g(x)=[x]f(x)=|x-1|, g(x)=[x] and h(x)=f(x)g(x)h(x)=f(x) g(x) where [.] is greatest integer function. What is ∫−10h(x)dx\int_{-1}^0 h(x) d x equal to?

Options

  1. A−32-\frac{3}{2}
  2. B-1
  3. C0
  4. D12\frac{1}{2}

Correct answer

A. −32-\frac{3}{2}

Step-by-step solution

 Given, f(x)=∣x−1∣...(i)g(x)=[x]...(ii)andh(x)=f(x)⋅g(x)⇒h(x)=h(x)−1∣⋅[x]...(iii)\begin{array}{ll} \text{ Given, } & f(x)=|x-1| \qquad ...(i) \\ & g(x)=[x] \qquad ...(ii) \\ and & h(x)=f(x) \cdot g(x) \\ \Rightarrow & h(x)=h(x)-1 \mid \cdot[x] \qquad ...(iii) \\ \end{array}∫−10h(x)dx=∫−10f(x)\cdotg(x)dx=∫−10∣x−1∣⋅[x]dx=∫−10(1−x)⋅(−1)dx=∫12(x−1)dx=[x22−x]12=[0−(12+1)]=−32\begin{aligned} \int_{-1}^0 h(x) d x & =\int_{-1}^0 f(x) \cdotg(x) d x \\ & =\int_{-1}^0|x-1| \cdot[x] d x \\ & =\int_{-1}^0(1-x) \cdot(-1) d x \\ & =\int_1^2(x-1) d x=\left[\frac{x^2}{2}-x\right]_1^2 \\ & =\left[0-\left(\frac{1}{2}+1\right)\right]=-\frac{3}{2}\end{aligned}$
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