If $[t]$ denotes the greatest integer $\leq t$, then the value of $\frac{3(\mathrm{e}-1)}{\mathrm{e}} \int_{1}^{2} \mathrm{x}^{2} \mathrm{e}^{[\mathrm{x}]+[\mathrm{x}^{3}]} \mathrm{dx}$ is:

(B) $\mathrm{e}^{8}-\mathrm{e}$

(A) $\mathrm{e}^{9}-\mathrm{e}$

JEE Main · 2023 · 30 Jan 2023 · Shift 1 · Definite Integrals

[t]

denotes the greatest integer

\leq t

, then the value of

\frac{3 ( e - 1 )}{e} \int_{1}^{2} x^{2} e^{[x] + [x^{3}]} dx

is:

Answer: (B)

On the interval

[1, 2)

the greatest integer

[x] = 1

, so the integral becomes

I = \int_{1}^{2} x^{2} e^{[x] + [x^{3}]} d x = e \int_{1}^{2} x^{2} e^{[x^{3}]} d x

. Substituting

t = x^{3}

d t = 3 x^{2} d x

, the limits run from

1

8

, giving

I = \frac{e}{3} \int_{1}^{8} e^{[t]} d t

. Splitting the range over the unit integer intervals,

\int_{1}^{8} e^{[t]} d t = e + e^{2} + \dots + e^{7} = \frac{e ( e ^{7} - 1 )}{e - 1}

, so

I = \frac{e ^{2} ( e ^{7} - 1 )}{3 ( e - 1 )}

. Multiplying by

\frac{3 ( e - 1 )}{e}

cancels the constants to leave

e (e^{7} - 1)

. The required value is

e^{8} - e

JEE Main · 2023 · 30 Jan 2023 · Shift 1 · Single correct option