JEE Main 2021 — Mathematics Calculus

Given $f(x)={e}^{-x}\mathrm{sin}x$f(x)=e−xsinx

Now, $\displaystyle F(x)={\int }_{0}^{x}f(t)dt$F(x)=∫0x​f(t)dt

Using Newton Leibnitz rule i.e. $\displaystyle \frac{d}{dx}{\int }_{u(x)}^{v(x)}f(t)dt=f(v(x))\cdot ({v}^{'}(x))-f(u(x))\cdot ({u}^{'}(x)),$dxd​∫u(x)v(x)​f(t)dt=f(v(x))⋅(v′(x))−f(u(x))⋅(u′(x)),

$$\Rightarrow {F}^{'}(x)=f(x)$$⇒F′(x)=f(x)

Now, $\displaystyle I={\int }_{0}^{1}({F}^{'}(x)+f(x)){e}^{x}\mathrm{dx}$I=∫01​(F′(x)+f(x))exdx

$$\Rightarrow I={\int }_{0}^{1}(f(x)+f(x))\cdot {e}^{x}\mathrm{dx}$$⇒I=∫01​(f(x)+f(x))⋅exdx

$$\Rightarrow I=2{\int }_{0}^{1}f(x)\cdot {e}^{x}dx$$⇒I=2∫01​f(x)⋅exdx

$$\Rightarrow I=2{\int }_{0}^{1}{e}^{-x}\mathrm{sin}x\cdot {e}^{x}dx$$⇒I=2∫01​e−xsinx⋅exdx

$$\Rightarrow I=2{\int }_{0}^{1}\mathrm{sin}xdx$$⇒I=2∫01​sinxdx

$$\Rightarrow I=2{[-\mathrm{cos}x]}_{0}^{1}$$⇒I=2[−cosx]01​

$$\Rightarrow I=2(1-\mathrm{cos}1)$$⇒I=2(1−cos1)

Using the expansion of $\mathrm{cos}x=1-\frac{{x}^{2}}{2!}+\frac{{x}^{4}}{4!}-\frac{{x}^{6}}{6!}+...$cosx=1−2!x2​+4!x4​−6!x6​+...

$$\Rightarrow I=2{1-(1-\frac{1}{2}+\frac{1}{4!}+\frac{1}{6!}+...)}$$⇒I=21−(1−21​+4!1​+6!1​+...)

$$\Rightarrow I=1-\frac{2}{4!}+\frac{2}{6!}-\frac{2}{9!}...$$⇒I=1−4!2​+6!2​−9!2​...

$$\Rightarrow 1-\frac{2}{4!}<I<1-\frac{2}{4!}+\frac{2}{6!}$$⇒1−4!2​<I<1−4!2​+6!2​

$$\Rightarrow \frac{11}{12}<I<\frac{331}{360}$$⇒1211​<I<360331​

$$\Rightarrow I\in [\frac{11}{12},\frac{331}{360}]$$⇒I∈[1211​,360331​]

$$\Rightarrow I\in [\frac{330}{360},\frac{331}{360}].$$⇒I∈[360330​,360331​].