Definite Integration Question 13
Question 13 - 30 January - Shift 1
If $[t]$ denotes the greatest integer $\leq 1$, then the value of $\frac{3(e-1)}{e} \int_1^{2} x^{2} e^{[x]+[x^{3}]} dx$ is :
(1) $e^{9}-e$
(2) $e^{8}-e$
(3) $e^{7}-1$
(4) $e^{8}-1$
Show Answer
Answer: (2)
Solution:
Formula: Integration by substitution, Properties of Greatest Integer Function, Sum of the terms in G.P
$ \begin{aligned} & I = \int_1^{2} x^{2} e^{[x^{3}]+1} dx \\ & \text{Substitute} \ x^{3}=t \Rightarrow 3 x^{2} d x=d t \\ & I=\frac{e}{3} \int_1^{8} e^{[t]} dt \\ & I=\frac{e}{3}{\int_1^{2} edt+\int_2^{3} e^{2} dt+\ldots \ldots \ldots+\int_7^{8} e^{7} dt} \\ & I=\frac{e}{3}(e+e^{2}+\ldots \ldots \ldots+e^{7}) \\ & I=\frac{e^{2}}{3}(1+e+\ldots \ldots . .+e^{6}) \\ & I=\frac{e^{2}}{3} \frac{(e^{7}-1)}{(e-1)} \\ & \text {Now,}\\ & \frac{3(e-1)}{e} \times I =\frac{3}{e}(e-1) \times \frac{e^{2}}{3} \frac{(e^{7}-1)}{(e-1)} \\ & \frac{3(e-1)}{e} \times I =e(e^{7}-1) \\ & \frac{3(e-1)}{e} \times I =e^{8}-e \\ & \frac{3(e-1)}{e} \int_1^{2} x^{2} e^{[x]+[x^{3}]} dx=e^{8}-e \end{aligned} $
