Definite Integration Question 20
Question 20
- Determine a positive integer $n \leq 5$, such that $\int_{0}^{1} e^{x}(x-1)^{n} d x=16-6 e$
$(1992,4 M)$
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Answer:
Correct Answer: 20. At $x=1$ and $\frac{7}{5}, f(x)$ is maximum and minimum respectively.
Solution:
- Let $I_{n}=\int_{0}^{1} e^{x}(x-1)^{n} d x$
$$ \text { Put } x-1=t \Rightarrow d x=d t $$
$\therefore \quad I_{n}=\int_{-1}^{0} e^{t+1} \cdot t^{n} d t=e \int_{-1}^{0} t^{n} e^{t} d t$
$=e\left[t^{n} e^{t}\right]{-1}^{0}-n \int{-1}^{0} t^{n-1} e^{t} d t$
$=e 0-(-1)^{n} e^{-1}-n \int_{-1}^{0} t^{n-1} e^{t} d t$
$=(-1)^{n+1}-n e \int_{-1}^{0} t^{n-1} e^{t} d t$
$\Rightarrow \quad I_{n}=(-1)^{n+1}-n I_{n-1}$
For $n=1, I_{1}=\int_{0}^{1} e^{x}(x-1) d x=\left[e^{x}(x-1)\right]{0}^{1}-\int{0}^{1} e^{x} d x$
$$ =e^{1}(1-1)-e^{0}(0-1)-\left[e^{x}\right]_{0}^{1}=1-(e-1)=2-e $$
Therefore, from Eq. (i), we get
$$ \text { and } \quad \begin{aligned} I_{2} & =(-1)^{2+1}-2 I_{1}=-1-2(2-e)=2 e-5 \ I_{3} & =(-1)^{3+1}-3 I_{2}=1-3(2 e-5)=16-6 e \end{aligned} $$
Hence, $n=3$ is the answer.
