Let f: R arrow R be defined as f(x)=e^{-x} x. If F:[0,1] arrow R is a differentiable function, such that F(x)=₀^{x} f(t) d t, then the value of ₀¹[F^{'}(x)+f(x)] e^{x} d x lies in the interval [17 Mar 2021 Shift 2]

Correct Answer is : [;330360,;331360]≤ft[\\;330/360,\\; 331/360][;360330,;360331]

Correct Answer is : [;330360,;331360]≤ft[\\;330/360,\\; 331/360][;360330,;360331]

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Definite Integrals Part 2

Section: Mathematics

Question : 87 of 100

Marks: +1, -0

Let

f: R \rightarrow R

f(x)=e^{-x} \sin x

F:[0,1] \rightarrow R

F(x)=\int\limits_{0}^{x} f(t) d t

\int\limits_{0}^{1}[F^{'}(x)+f(x)] e^{x} d x

Solution:

Given,

f(x)=e^{-x} \cdot \sin x

and

F(x)=\int\limits_{0}^{x} f(t) d t

\because F(x)

is differentiable function.

\therefore F^{'}(x)=f(x) \times 1-f(0) \times 0

(using Newton-Leibnitz rule)

\Rightarrow \\;\\; F^{'}(x)=f(x)

...(i)

Let

I=\int\limits_{0}^{1}[F^{'}(x)+f(x)] e^{x} d x

=\int\limits_{0}^{1}[f(x)+f(x)] e^{x} d x=\int\limits_{0}^{1} 2 \cdot f(x) \cdot e^{x} d x

[ from Eq. (i) ]

\Rightarrow \\;\\; I=2 \cdot \int\limits_{0}^{1} f(x) \cdot e^{x} d x

=2 \cdot \int\limits_{0}^{1} e^{-x} \sin x \cdot e^{x} d x

=2 \int\limits_{0}^{1} \sin x d x=2[-\cos x]_{0}^{1}

=2[-(\cos 1-\cos 0)]=2(1-\cos 1)

\Rightarrow l=2 \cdot \left[1-\left(1-\\;\frac{(1)^{2}}{2!}+\\;\frac{(1)^{4}}{4!}-\\;\frac{(1)^{6}}{6!}+\\;\frac{(1)^{8}}{8!}-\dots\right)\right]

[using expansion of

\cos x

i.e.,

\cos x=1-\\;\frac{x^{2}}{2!}+\\;\frac{x^{4}}{4!}-\\;\frac{x^{6}}{6!}+\\;\frac{x^{8}}{8!}-\dots 1

I=2\left[1-1+\\;\frac{1}{2!}-\\;\frac{1}{4!}+\\;\frac{1}{6!}-\\;\frac{1}{8!}+\dots\right]

\Rightarrow \\;\\; I=2\left(\\;\frac{1}{2}-\\;\frac{1}{24}+\\;\frac{1}{720}-\dots\right)

Now,

I< 2\left(\\;\frac{1}{2}-\\;\frac{1}{24}+\\;\frac{1}{720}\right)

\Rightarrow \\;\\; I< \left(1-\\;\frac{1}{12}+\\;\frac{1}{360}\right) \Rightarrow I< \\;\frac{360-30+1}{360}

\Rightarrow \\;\\; I< \\;\frac{331}{360}

...(ii)

Also,

I>2\left(\\;\frac{1}{2}-\\;\frac{1}{24}\right) \Rightarrow I>\left(1-\\;\frac{1}{12}\right)

\Rightarrow \\;\\; I>\\;\frac{11}{12} \Rightarrow \\;\\; I>\\;\frac{11 \times 30}{12 \times 30}

\Rightarrow \\;\\; I>\\;\frac{330}{360}

....(iii)

From Eqs. (ii) and (iii), we get

\\;\frac{330}{360} < l < \\;\frac{331}{360}

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