MHT CET · Maths · Definite Integration
The value of integral \(\int_{-2}^0\left(x^3+3 x^2+3 x+5+(x+1) \cos (x+1)\right) d x\) is equal to
- A 0
- B 6
- C 4
- D 8
Answer & Solution
Correct Answer
(D) 8
Step-by-step Solution
Detailed explanation
Let
\(I =\int_{-2}^0[x^3+3 x^2+3 x+5+(x+1) \) \(\cos (x+1)] d x \)
\( =\int_{-2}^0[(x+1)^3+4+(x+1)\) \(\cos (x+1)] d x\)
\(\text {Put } x+1=t \Rightarrow d x=d t \)
\( \therefore I=\int_{-1}^1\left(t^3+4+t \cos t\right) d t\)
Since \(t^3\) and \(t \cos t\) are odd functions.
\(\therefore I=\int_{-1}^1 4 d t=4[t]_{-1}^1=8\)
[Note: The answer of the question is not mentioned as an option.]
\(I =\int_{-2}^0[x^3+3 x^2+3 x+5+(x+1) \) \(\cos (x+1)] d x \)
\( =\int_{-2}^0[(x+1)^3+4+(x+1)\) \(\cos (x+1)] d x\)
\(\text {Put } x+1=t \Rightarrow d x=d t \)
\( \therefore I=\int_{-1}^1\left(t^3+4+t \cos t\right) d t\)
Since \(t^3\) and \(t \cos t\) are odd functions.
\(\therefore I=\int_{-1}^1 4 d t=4[t]_{-1}^1=8\)
[Note: The answer of the question is not mentioned as an option.]
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