JEE Main 2023 — Mathematics Calculus

Let

$$I={\int }_{1}^{2}{x}^{2}{e}^{[x]+[{x}^{3}]}dx$$I=∫12​x2e[x]+[x3]dx

$$\Rightarrow I={\int }_{1}^{2}{x}^{2}{e}^{[{x}^{3}]+1}dx$$⇒I=∫12​x2e[x3]+1dx

$$\Rightarrow I=e{\int }_{1}^{2}{x}^{2}{e}^{[{x}^{3}]}dx$$⇒I=e∫12​x2e[x3]dx

Put ${x}^{3}=t$x3=t

$$\Rightarrow 3{x}^{2}dx=dt$$⇒3x2dx=dt

So,

$$I=\frac{e}{3}{\int }_{1}^{8}{e}^{[t]}dt$$I=3e​∫18​e[t]dt

$$\Rightarrow I=\frac{e}{3}{{\int }_{1}^{2}edt+{\int }_{2}^{3}{e}^{2}dt+\ldots \ldots \ldots +{\int }_{7}^{8}{e}^{7}dt}$$⇒I=3e​∫12​edt+∫23​e2dt+………+∫78​e7dt

$$\Rightarrow I=\frac{e}{3}(e+{e}^{2}+\ldots \ldots ..+{e}^{7})$$⇒I=3e​(e+e2+……..+e7)

$$\Rightarrow I=\frac{{e}^{2}}{3}\frac{({e}^{7}-1)}{(e-1)}$$⇒I=3e2​(e−1)(e7−1)​

So,

$$\frac{3(e-1{)}^{2}}{e}{\int }_{1}^{2}{x}^{2}{e}^{[x]+[{x}^{3}]}dx=\frac{3}{e}(e-1)\times \frac{{e}^{2}}{3}\frac{({e}^{7}-1)}{(e-1)}$$e3(e−1)2​∫12​x2e[x]+[x3]dx=e3​(e−1)×3e2​(e−1)(e7−1)​

$$\Rightarrow \frac{3(e-1{)}^{2}}{e}{\int }_{1}^{2}{x}^{2}{e}^{[x]+[{x}^{3}]}dx={e}^{8}-e$$⇒e3(e−1)2​∫12​x2e[x]+[x3]dx=e8−e