Question:

If $[ t ]$ denotes the greatest integer $\leq t$, then the value of $\frac{3( e -1)}{ e } \int\limits_1^2 x^2 e^{[x]+\left[x^3\right]} dx$ is :

Updated On: Mar 20, 2025

$e^8-1$
$e^7-1$
$e^9-e$
$e^8-e$

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The Correct Option is D

Solution and Explanation

1 \int 2 x^{2} e^{[x^{3}] + 1} d x

x^{3} = t

3 x^{2} d x = d t

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

= \frac{e}{3} {1 \int 2 e d t + \int_{2}^{3} e^{2} d t + \dots\dots .. + 7 \int 8 e^{7} d t}

= \frac{e}{3} (e + e^{2} + \dots\dots\dots + e^{7})

= \frac{e ^{2}}{3} (1 + e + \dots\dots\dots + e^{6}) = \frac{e ^{2}}{3} \frac{( e ^{7} - 1 )}{( e - 1 )}

\frac{3 ( e - 1 )}{e} 1 \int 2 x^{2} \times e^{[x] + [x^{3}]} d x = \frac{3}{e} (e - 1) \times \frac{e ^{2}}{3} \frac{( e ^{7} - 1 )}{( e - 1 )}

= e (e^{7} - 1)

= e^{8} - e

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Concepts Used:

Application of Derivatives

Various Applications of Derivatives-

Rate of Change of Quantities:

If some other quantity ‘y’ causes some change in a quantity of surely ‘x’, in view of the fact that an equation of the form y = f(x) gets consistently pleased, i.e, ‘y’ is a function of ‘x’ then the rate of change of ‘y’ related to ‘x’ is to be given by

$\frac{\triangle y}{\triangle x}=\frac{y_2-y_1}{x_2-x_1}$

This is also known to be as the Average Rate of Change.

Increasing and Decreasing Function:

Consider y = f(x) be a differentiable function (whose derivative exists at all points in the domain) in an interval x = (a,b).

If for any two points x₁ and x₂ in the interval x such a manner that x₁ < x₂, there holds an inequality f(x₁) ≤ f(x₂); then the function f(x) is known as increasing in this interval.
Likewise, if for any two points x₁ and x₂ in the interval x such a manner that x₁ < x₂, there holds an inequality f(x₁) ≥ f(x₂); then the function f(x) is known as decreasing in this interval.
The functions are commonly known as strictly increasing or decreasing functions, given the inequalities are strict: f(x₁) < f(x₂) for strictly increasing and f(x₁) > f(x₂) for strictly decreasing.

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