SATHEE: Definite Integration Question 23

Definite Integration Question 23

Question 23

The value of

$\int_{- 2}^{0} [x^{3} + 3 x^{2} + 3 x + 3 + (x + 1) \cos (x + 1)] d x$ is $(2005, 1 M)$ (a) 0 (b) 3 (c) 4 (d) 1

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Solution:

Let $I = \int_{- 2}^{0} [x^{3} + 3 x^{2} + 3 x + 3 + (x + 1) \cos (x + 1)] d x$

$= \int_{- 2}^{0} [(x + 1)^{3} + 2 + (x + 1) \cos (x + 1)] d x$

Put $x + 1 = t$

$\begin{aligned} \Rightarrow d x = d t & ∴ I = \int_{- 1}^{1} (t^{3} + 2 + t \cos t) d t & = \int_{- 1}^{1} t^{3} d t + 2 \int_{- 1}^{1} d t + \int_{- 1}^{1} t \cos t d t & = 0 + 2 \cdot 2 [x]_{0}^{1} + 0 & = 4 \end{aligned}$

[since, $t^{3}$ and $t \cos t$ are odd functions]

Definite Integration Question 23

Question 23

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Definite Integration Question 23

Question 23

Engineering Preparation

Medical Preparation

NCERT Books & Solutions

Important Resources

Other Resources

Syllabus

Test Platform

Sample Mock Test

Mentorship

NCERT Exercise Video Solution

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