Let f(x) be a differentiable function defined on [0, 2] such that f’(x) = f’(2 – x) for all <math> <mrow> <mi>x</mi> <mo>∈</mo> <mrow> <mo>(</mo> <mrow> <mn>0</mn> <mo>,</mo> <mn>2</mn> </mrow> <mo>)</mo> </mrow> <mo>,</mo> </mrow> </math> f(0) = 1 and f(2) = e2. Then the value of <math> <mrow> <mstyle displaystyle="true"> <mrow> <munderover> <mo>∫</mo> <mrow> <mn>0</mn> </mrow> <mrow> <mn>2</mn> </mrow> </munderover> <mrow> <mi>f</mi> <mrow> <mo>(</mo> <mrow> <mi>x</mi> </mrow> <mo>)</mo> </mrow> <mi>d</mi> <mi>x</mi> </mrow> </mrow> </mstyle> <mtext> </mtext> <mtext> </mtext> <mi>i</mi> <mi>s</mi> </mrow> </math>

l = ∫ 0 2 f ( x ) d x = [ x f ( x ) ] 0 2 − ∫ 0 2 x f ' ( x ) d x = 2 e 2 − ∫ 0 2 x f ' ( x ) d x .(A)Put l 1 = ∫ 0 2 x f ' ( x ) d x .(i)Using properties ∫ a b f ( x ) d x = ∫ a b f ( a + b − x ) d x  l 1 = ∫ 0 2 ( 2 − x ) f ' ( 2 − x ) d x = ∫ 0 2 ( 2 − x ) f ' ( x ) d x .(ii)Adding (i) and (ii) we get 2 l 1 = 2 ∫ 0 2 f ' ( x ) d x ⇒ l 1 = [ f ( x ) ] 0 2 f(2) – f(0) = e2 – 1From (A) l = 2e2 – e2 + 1 = e2 + 1

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Definite Integrals Maths Integrals Maths Class 12th

**Let f(x) be a differentiable function defined on [0, 2] such that f’(x) = f’(2 – x) for all $x \in (0, 2),$ f(0) = 1 and f(2) = e². Then the value of $\int_{0}^{2} f (x) d x i s$**

Option 1 -

2 (1 + e²)

Option 2 -

1 – e²

Option 3 -

1 + e²

Option 4 -

2 (1 – e²)

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1 Answer

Answered by

Vishal Baghel | Contributor-Level 10

a month ago

Correct Option - 3

Detailed Solution:

$l = \int_{0}^{2} f (x) d x = {[x f (x)]}_{0}^{2} - \int_{0}^{2} x f' (x) d x = 2 e^{2} - \int_{0}^{2} x f' (x) d x$ .(A)

Put $l_{1} = \int_{0}^{2} x f' (x) d x$ .(i)

Using properties $\int_{a}^{b} f (x) d x = \int_{a}^{b} f (a + b - x) d x$

$l_{1} = \int_{0}^{2} (2 - x) f' (2 - x) d x = \int_{0}^{2} (2 - x) f' (x) d x$ .(ii)

Adding (i) and (ii) we get

$2 l_{1} = 2 \int_{0}^{2} f' (x) d x \Rightarrow l_{1} = {[f (x)]}_{0}^{2}$

f(2) – f(0) = e² – 1

From (A) l = 2e² – e² + 1 = e² + 1

<math> <mrow> <mi>l</mi> <mo>=</mo> <mstyle displaystyle="true"> <mrow> <munderover> <mo>∫</mo> <mrow> <mn>0</mn> </mrow> <mrow> <mn>2</mn> </mrow> </munderover> <mrow> <mi>f</mi> <mrow> <mo>(</mo> <mrow> <mi>x</mi> </mrow> <mo>)</mo> </mrow> <mi>d</mi> <mi>x</mi> <mo>=</mo> <msubsup> <mrow> <mrow> <mo>[</mo> <mrow> <mi>x</mi> <mi>f</mi> <mrow> <mo>(</mo> <mrow> <mi>x</mi> </mrow> <mo>)</mo> </mrow> </mrow> <mo>]</mo> </mrow> </mrow> <mrow> <mn>0</mn> </mrow> <mrow> <mn>2</mn> </mrow> </msubsup> <mo>−</mo> <mstyle displaystyle="true"> <mrow> <munderover> <mo>∫</mo> <mrow> <mn>0</mn> </mrow> <mrow> <mn>2</mn> </mrow> </munderover> <mrow> <mi>x</mi> <mi>f</mi> <mo>'</mo> <mrow> <mo>(</mo> <mrow> <mi>x</mi> </mrow> <mo>)</mo> </mrow> <mi>d</mi> <mi>x</mi> <mo>=</mo> <mn>2</mn> <msup> <mrow> <mi>e</mi> </mrow> <mrow> <mn>2</mn> </mrow> </msup> <mo>−</mo> <mstyle displaystyle="true"> <mrow> <munderover> <mo>∫</mo> <mrow> <mn>0</mn> </mrow> <mrow> <mn>2</mn> </mrow> </munderover> <mrow> <mi>x</mi> <mi>f</mi> <mo>'</mo> <mrow> <mo>(</mo> <mrow> <mi>x</mi> </mrow> <mo>)</mo> </mrow> <mi>d</mi> <mi>x</mi> </mrow> </mrow> </mstyle> </mrow> </mrow> </mstyle> </mrow> </mrow> </mstyle> </mrow> </math>    .(A)Put <math> <mrow> <msub> <mrow> <mi>l</mi> </mrow> <mrow> <mn>1</mn> </mrow> </msub> <mo>=</mo> <mstyle displaystyle="true"> <mrow> <munderover> <mo>∫</mo> <mrow> <mn>0</mn> </mrow> <mrow> <mn>2</mn> </mrow> </munderover> <mrow> <mi>x</mi> <mi>f</mi> <mo>'</mo> <mrow> <mo>(</mo> <mrow> <mi>x</mi> </mrow> <mo>)</mo> </mrow> <mi>d</mi> <mi>x</mi> </mrow> </mrow> </mstyle> </mrow> </math>               .(i)Using properties <math> <mrow> <mstyle displaystyle="true"> <mrow> <munderover> <mo>∫</mo> <mrow> <mi>a</mi> </mrow> <mrow> <mi>b</mi> </mrow> </munderover> <mrow> <mi>f</mi> <mrow> <mo>(</mo> <mrow> <mi>x</mi> </mrow> <mo>)</mo> </mrow> <mi>d</mi> <mi>x</mi> <mo>=</mo> <mstyle displaystyle="true"> <mrow> <munderover> <mo>∫</mo> <mrow> <mi>a</mi> </mrow> <mrow> <mi>b</mi> </mrow> </munderover> <mrow> <mi>f</mi> <mrow> <mo>(</mo> <mrow> <mi>a</mi> <mo>+</mo> <mi>b</mi> <mo>−</mo> <mi>x</mi> </mrow> <mo>)</mo> </mrow> <mi>d</mi> <mi>x</mi> </mrow> </mrow> </mstyle> </mrow> </mrow> </mstyle> </mrow> </math>  <math> <mrow> <msub> <mrow> <mi>l</mi> </mrow> <mrow> <mn>1</mn> </mrow> </msub> <mo>=</mo> <mstyle displaystyle="true"> <mrow> <munderover> <mo>∫</mo> <mrow> <mn>0</mn> </mrow> <mrow> <mn>2</mn> </mrow> </munderover> <mrow> <mrow> <mo>(</mo> <mrow> <mn>2</mn> <mo>−</mo> <mi>x</mi> </mrow> <mo>)</mo> </mrow> <mi>f</mi> <mo>'</mo> <mrow> <mo>(</mo> <mrow> <mn>2</mn> <mo>−</mo> <mi>x</mi> </mrow> <mo>)</mo> </mrow> <mi>d</mi> <mi>x</mi> <mo>=</mo> <mstyle displaystyle="true"> <mrow> <munderover> <mo>∫</mo> <mrow> <mn>0</mn> </mrow> <mrow> <mn>2</mn> </mrow> </munderover> <mrow> <mrow> <mo>(</mo> <mrow> <mn>2</mn> <mo>−</mo> <mi>x</mi> </mrow> <mo>)</mo> </mrow> <mi>f</mi> <mo>'</mo> <mrow> <mo>(</mo> <mrow> <mi>x</mi> </mrow> <mo>)</mo> </mrow> <mi>d</mi> <mi>x</mi> </mrow> </mrow> </mstyle> </mrow> </mrow> </mstyle> </mrow> </math> .(ii)Adding (i) and (ii) we get <math> <mrow> <mn>2</mn> <msub> <mrow> <mi>l</mi> </mrow> <mrow> <mn>1</mn> </mrow> </msub> <mo>=</mo> <mn>2</mn> <mstyle displaystyle="true"> <mrow> <munderover> <mo>∫</mo> <mrow> <mn>0</mn> </mrow> <mrow> <mn>2</mn> </mrow> </munderover> <mrow> <mi>f</mi> <mo>'</mo> <mrow> <mo>(</mo> <mrow> <mi>x</mi> </mrow> <mo>)</mo> </mrow> <mi>d</mi> <mi>x</mi> <mo>⇒</mo> <msub> <mrow> <mi>l</mi> </mrow> <mrow> <mn>1</mn> </mrow> </msub> <mo>=</mo> <msubsup> <mrow> <mrow> <mo>[</mo> <mrow> <mi>f</mi> <mrow> <mo>(</mo> <mrow> <mi>x</mi> </mrow> <mo>)</mo> </mrow> </mrow> <mo>]</mo> </mrow> </mrow> <mrow> <mn>0</mn> </mrow> <mrow> <mn>2</mn> </mrow> </msubsup> </mrow> </mrow> </mstyle> </mrow> </math> f(2) – f(0) = e2 – 1From (A) l = 2e2 – e2 + 1 = e2 + 1

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**Let f(x) be a differentiable function defined on [0, 2] such that f’(x) = f’(2 – x) for all $x \in (0, 2),$ f(0) = 1 and f(2) = e². Then the value of $\int_{0}^{2} f (x) d x i s$**

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Let f(x) be a differentiable function defined on [0, 2] such that f’(x) = f’(2 – x) for all x ∈ ( 0 , 2 ) , f(0) = 1 and f(2) = e2. Then the value of ∫ 0 2 f ( x ) d x i s

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**Let f(x) be a differentiable function defined on [0, 2] such that f’(x) = f’(2 – x) for all $x \in (0, 2),$ f(0) = 1 and f(2) = e². Then the value of $\int_{0}^{2} f (x) d x i s$**