KCET · Maths · Definite Integration
\(\int_{-2}^0\left(x^3+3 x^2+3 x+3+(x+1) \cos (x+1)\right) d x\)
is equals to
- A \(3\)
- B \(4\)
- C \(1\)
- D \(0\)
Answer & Solution
Correct Answer
(B) \(4\)
Step-by-step Solution
Detailed explanation
Let \(\begin{aligned} I & =\int_{-2}^0 x^3+3 x^2+3 x+3+(x+1) \cos (x+1) d x \\ I & =\int_{-2}^0\left[(x+1)^3+2+(x+1) \cos (x+1) d x\right.\end{aligned}\)
Putting, \(x+1=t\).
Then \(d x=d t\), we get
\(\begin{aligned} & I=\int_{-1}^1\left[t^3+2+t \cos t\right] d t \\ & I=\int_{-1}^1(2) \text { u. }=[2 t]_{-1}^1=2[1-(-1)]=4\end{aligned}\)
As \(t^3+t \cos t\) is an odd function.
Putting, \(x+1=t\).
Then \(d x=d t\), we get
\(\begin{aligned} & I=\int_{-1}^1\left[t^3+2+t \cos t\right] d t \\ & I=\int_{-1}^1(2) \text { u. }=[2 t]_{-1}^1=2[1-(-1)]=4\end{aligned}\)
As \(t^3+t \cos t\) is an odd function.
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