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NDA 2024
Exam
NDA
Year
2024
Subject
Mathematics
Chapter
Definite Integration

NDA 2024 · Mathematics · Definite Integration

Paragraph: Let f(x)=|x^2-x-2| Question: What is ∫_1^3 f(x) d x...

Paragraph: Let f(x)=∣x2−x−2∣f(x)=\left|x^2-x-2\right|
Question: What is ∫13f(x)dx\int_1^3 f(x) d x equal to?

Options

  1. A2
  2. B3
  3. C4
  4. D5

Correct answer

B. 3

Step-by-step solution

f(x)=∣x2−x−2∣{x2−x−2;x∈(−∞,−1]∪[2,∞)−(x2−x−2);x∈[−1,2]Let I=∫13f(x)dn=−∫12(x2−x−2)dx+∫23(x2−x−2)dx=−[x33−x22−2x]12+[x33−x22−2x]23=−[83−42−4]+[13−12−2]+[273−92−6]−[83−42−4]=206−136−96+206=186=3\begin{aligned} & f(x)=\left|x^2-x-2\right| \\ & \qquad\left\{\begin{array}{l}x^2-x-2 ; x \in(-\infty,-1] \cup[2, \infty) \\ -\left(x^2-x-2\right) ; x \in[-1,2]\end{array}\right. \\ & \text{Let } I=\int_1^3 f(x) d n=-\int_1^2\left(x^2-x-2\right) d x+\int_2^3\left(x^2-x-2\right) d x \\ & \quad=-\left[\frac{x^3}{3}-\frac{x^2}{2}-2 x\right]_1^2+\left[\frac{x^3}{3}-\frac{x^2}{2}-2 x\right]_2^3 \\ & =-\left[\frac{8}{3}-\frac{4}{2}-4\right]+\left[\frac{1}{3}-\frac{1}{2}-2\right]+\left[\frac{27}{3}-\frac{9}{2}-6\right]-\left[\frac{8}{3}-\frac{4}{2}-4\right] \\ & \quad=\frac{20}{6}-\frac{13}{6}-\frac{9}{6}+\frac{20}{6}=\frac{18}{6}=3\end{aligned}
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