SATHEE: Definite Integration Question 20

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 $I_{n} = \int_{0}^{1} e^{x} (x - 1)^{n} d x$

$Put x - 1 = t \Rightarrow d x = d t$

$∴ I_{n} = \int_{- 1}^{0} e^{t + 1} \cdot t^{n} d t = e \int_{- 1}^{0} t^{n} e^{t} d t$

$=e\left[t^{n} e^{t}\right]{-1}^{0}-n \int{-1}^{0} t^{n-1} e^{t} d t$

$= e 0 - (- 1)^{n} e^{- 1} - n \int_{- 1}^{0} t^{n - 1} e^{t} d t$

$= (- 1)^{n + 1} - n e \int_{- 1}^{0} t^{n - 1} e^{t} d t$

$\Rightarrow I_{n} = (- 1)^{n + 1} - n I_{n - 1}$

For $n=1, I_{1}=\int_{0}^{1} e^{x}(x-1) d x=\left[e^{x}(x-1)\right]{0}^{1}-\int{0}^{1} e^{x} d x$

$= e^{1} (1 - 1) - e^{0} (0 - 1) - {[e^{x}]}_{0}^{1} = 1 - (e - 1) = 2 - e$

Therefore, from Eq. (i), we get

$and \begin{aligned} I_{2} & = (- 1)^{2 + 1} - 2 I_{1} = - 1 - 2 (2 - e) = 2 e - 5 I_{3} & = (- 1)^{3 + 1} - 3 I_{2} = 1 - 3 (2 e - 5) = 16 - 6 e \end{aligned}$

Hence, $n = 3$ is the answer.

Definite Integration Question 20

Question 20

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Definite Integration Question 20

Question 20

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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