JEE Mains · Maths · STD 12 - 7.2 definite integral
Let \(f(x)=\max \{|x+1|,|x+2|, \ldots,|x+5|\}\). Then \(\int_{-6}^{0} f(x) d x\) is equal to
- A \(20\)
- B \(40\)
- C \(21\)
- D \(41\)
Answer & Solution
Correct Answer
(C) \(21\)
Step-by-step Solution
Detailed explanation
\(f(x)=\max \{|x+1|,|x+2|,|x+3|,|x+4|,|x+5|\}\) \(\int_{-6}^{0} f(x) d x=\int_{-6}^{-3}|x+1| d x+\int_{-3}^{0}|x+5| d x\) \(=-\int_{-6}^{-3}(x+1) d x+\int_{-3}^{0}(x+5) d x\) \(=-\left[\frac{x^{2}}{2}+x\right]_{-6}^{-3}+\left[\frac{x^{2}}{2}+5 x\right]_{-3}^{0}\)…
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