JEE Main 2019 — Mathematics Calculus

Let $\displaystyle I=\int {x}^{5}{e}^{-{x}^{2}}dx$I=∫x5e−x2dx

Let, $-{x}^{2}=t\Rightarrow xdx=-\frac{1}{2}dt$−x2=t⇒xdx=−21​dt

$$\therefore I=-\frac{1}{2}\int {t}^{2}{e}^{t}dt$$∴I=−21​∫t2etdt

By using integrating by parts, i.e. $\displaystyle \int (u\cdot v)dx=u\int vdx-\int [\frac{d}{dx}u\int vdx]dx+c,$∫(u⋅v)dx=u∫vdx−∫[dxd​u∫vdx]dx+c, we get

$$I=-\frac{1}{2}[{t}^{2}\int {e}^{t}dt-\int (2t\int {e}^{t}dt)dt]+c$$I=−21​[t2∫etdt−∫(2t∫etdt)dt]+c

$$\Rightarrow I=-\frac{1}{2}[{t}^{2}{e}^{t}-2\int t{e}^{t}dt]+c$$⇒I=−21​[t2et−2∫tetdt]+c

Again, applying integrating by parts, we get

$$\Rightarrow I=-\frac{1}{2}[{t}^{2}{e}^{t}-2(t\int {e}^{t}dt-\int 1\cdot {e}^{t}dt)]+c$$⇒I=−21​[t2et−2(t∫etdt−∫1⋅etdt)]+c

$$\Rightarrow I=-\frac{1}{2}[{t}^{2}{e}^{t}-2(t{e}^{t}-{e}^{t})]+c$$⇒I=−21​[t2et−2(tet−et)]+c

$$\Rightarrow I=-\frac{1}{2}[{t}^{2}{e}^{t}-2t{e}^{t}+2{e}^{t}]+c$$⇒I=−21​[t2et−2tet+2et]+c

$$\Rightarrow I=-\frac{{e}^{t}}{2}[{t}^{2}-2t+2]+c$$⇒I=−2et​[t2−2t+2]+c

$$\Rightarrow I=-\frac{{e}^{-{x}^{2}}}{2}[{x}^{4}+2{x}^{2}+2]+c$$⇒I=−2e−x2​[x4+2x2+2]+c

Given, $\displaystyle I=\int {x}^{5}{e}^{-{x}^{2}}dx=g(x){e}^{-{x}^{2}}+c$I=∫x5e−x2dx=g(x)e−x2+c

$$\Rightarrow g(x)=-\frac{1}{2}({x}^{4}+2{x}^{2}+2)$$⇒g(x)=−21​(x4+2x2+2)

$$\Rightarrow g(-1)=-\frac{1}{2}({(-1)}^{4}+2{(-1)}^{2}+2)$$⇒g(−1)=−21​((−1)4+2(−1)2+2)

$$\Rightarrow g(-1)=-\frac{5}{2}.$$⇒g(−1)=−25​.