Accumulation Function Analysis

Given a graph or definition of a function f, analyze the function g(x) = ∫ₐˣ f(t)dt for properties such as increasing/decreasing intervals, extrema, inflection points, concavity, or specific values.

ap-calculus-bc 2006 Q4 View
The continuous function $f$ is defined on the interval $- 4 \leq x \leq 3$. The graph of $f$ consists of two quarter circles and one line segment, as shown in the figure above. Let $g ( x ) = 2 x + \int _ { 0 } ^ { x } f ( t ) d t$. (a) Find $g ( - 3 )$. Find $g ^ { \prime } ( x )$ and evaluate $g ^ { \prime } ( - 3 )$. (b) Determine the $x$-coordinate of the point at which $g$ has an absolute maximum on the interval $- 4 \leq x \leq 3$. Justify your answer. (c) Find all values of $x$ on the interval $- 4 < x < 3$ for which the graph of $g$ has a point of inflection. Give a reason for your answer. (d) Find the average rate of change of $f$ on the interval $- 4 \leq x \leq 3$. There is no point $c , - 4 < c < 3$, for which $f ^ { \prime } ( c )$ is equal to that average rate of change. Explain why this statement does not contradict the Mean Value Theorem.
ap-calculus-bc 2008 Q3 View
Let $f$ be the continuous function defined on $[ - 4,3 ]$ whose graph, consisting of three line segments and a semicircle centered at the origin, is given above. Let $g$ be the function given by $g ( x ) = \int _ { 1 } ^ { x } f ( t ) d t$.
(a) Find the values of $g ( 2 )$ and $g ( - 2 )$.
(b) For each of $g ^ { \prime } ( - 3 )$ and $g ^ { \prime \prime } ( - 3 )$, find the value or state that it does not exist.
(c) Find the $x$-coordinate of each point at which the graph of $g$ has a horizontal tangent line. For each point, determine whether it is a relative maximum, a relative minimum, or neither.
(d) For $- 4 < x < 3$, find all values of $x$ for which the graph of $g$ has a point of inflection. Explain your reasoning.
ap-calculus-bc 2009 Q3 View
A continuous function $f$ is defined on the closed interval $- 4 \leq x \leq 6$. The graph of $f$ consists of a line segment and a curve that is tangent to the $x$-axis at $x = 3$, as shown in the figure above. On the interval $0 < x < 6$, the function $f$ is twice differentiable, with $f ^ { \prime \prime } ( x ) > 0$.
(a) Is $f$ differentiable at $x = 0$ ? Use the definition of the derivative with one-sided limits to justify your answer.
(b) For how many values of $a , - 4 \leq a < 6$, is the average rate of change of $f$ on the interval $[ a , 6 ]$ equal to 0 ? Give a reason for your answer.
(c) Is there a value of $a , - 4 \leq a < 6$, for which the Mean Value Theorem, applied to the interval $[ a , 6 ]$, guarantees a value $c , a < c < 6$, at which $f ^ { \prime } ( c ) = \frac { 1 } { 3 }$ ? Justify your answer.
(d) The function $g$ is defined by $g ( x ) = \int _ { 0 } ^ { x } f ( t ) d t$ for $- 4 \leq x \leq 6$. On what intervals contained in $[ - 4,6 ]$ is the graph of $g$ concave up? Explain your reasoning.
ap-calculus-bc 2011 Q4 View
The continuous function $f$ is defined on the interval $-4 \leq x \leq 3$. The graph of $f$ consists of two quarter circles and one line segment, as shown in the figure above. Let $g(x) = 2x + \int_{0}^{x} f(t)\, dt$.
(a) Find $g(-3)$. Find $g'(x)$ and evaluate $g'(-3)$.
(b) Determine the $x$-coordinate of the point at which $g$ has an absolute maximum on the interval $-4 \leq x \leq 3$. Justify your answer.
(c) Find all values of $x$ on the interval $-4 < x < 3$ for which the graph of $g$ has a point of inflection. Give a reason for your answer.
(d) Find the average rate of change of $f$ on the interval $-4 \leq x \leq 3$. There is no point $c$, $-4 < c < 3$, for which $f'(c)$ is equal to that average rate of change. Explain why this statement does not contradict the Mean Value Theorem.
ap-calculus-bc 2011 Q4 View
The graph of the differentiable function $y = f ( x )$ with domain $0 \leq x \leq 10$ is shown in the figure above. The area of the region enclosed between the graph of $f$ and the $x$-axis for $0 \leq x \leq 5$ is 10 , and the area of the region enclosed between the graph of $f$ and the $x$-axis for $5 \leq x \leq 10$ is 27 . The arc length for the portion of the graph of $f$ between $x = 0$ and $x = 5$ is 11, and the arc length for the portion of the graph of $f$ between $x = 5$ and $x = 10$ is 18 . The function $f$ has exactly two critical points that are located at $x = 3$ and $x = 8$.
(a) Find the average value of $f$ on the interval $0 \leq x \leq 5$.
(b) Evaluate $\int _ { 0 } ^ { 10 } ( 3 f ( x ) + 2 ) d x$. Show the computations that lead to your answer.
(c) Let $g ( x ) = \int _ { 5 } ^ { x } f ( t ) d t$. On what intervals, if any, is the graph of $g$ both concave up and decreasing? Explain your reasoning.
(d) The function $h$ is defined by $h ( x ) = 2 f \left( \frac { x } { 2 } \right)$. The derivative of $h$ is $h ^ { \prime } ( x ) = f ^ { \prime } \left( \frac { x } { 2 } \right)$. Find the arc length of the graph of $y = h ( x )$ from $x = 0$ to $x = 20$.
ap-calculus-bc 2012 Q3 View
Let $f$ be the continuous function defined on $[-4, 3]$ whose graph, consisting of three line segments and a semicircle centered at the origin, is given above. Let $g$ be the function given by $g(x) = \int_{1}^{x} f(t)\, dt$.
(a) Find the values of $g(2)$ and $g(-2)$.
(b) For each of $g'(-3)$ and $g''(-3)$, find the value or state that it does not exist.
(c) Find the $x$-coordinate of each point at which the graph of $g$ has a horizontal tangent line. For each of these points, determine whether $g$ has a relative minimum, relative maximum, or neither a minimum nor a maximum at the point. Justify your answers.
(d) For $-4 < x < 3$, find all values of $x$ for which the graph of $g$ has a point of inflection. Explain your reasoning.
ap-calculus-bc 2012 Q15 View
The graph of a differentiable function $f$ is shown above. If $h ( x ) = \int _ { 0 } ^ { x } f ( t ) d t$, which of the following is true?
(A) $h ( 6 ) < h ^ { \prime } ( 6 ) < h ^ { \prime \prime } ( 6 )$
(B) $h ( 6 ) < h ^ { \prime \prime } ( 6 ) < h ^ { \prime } ( 6 )$
(C) $h ^ { \prime } ( 6 ) < h ( 6 ) < h ^ { \prime \prime } ( 6 )$
(D) $h ^ { \prime \prime } ( 6 ) < h ( 6 ) < h ^ { \prime } ( 6 )$
(E) $h ^ { \prime \prime } ( 6 ) < h ^ { \prime } ( 6 ) < h ( 6 )$
ap-calculus-bc 2012 Q3 (Free Response) View
Let $f$ be the continuous function defined on $[ - 4,3 ]$ whose graph, consisting of three line segments and a semicircle centered at the origin, is given above. Let $g$ be the function given by $g ( x ) = \int _ { 1 } ^ { x } f ( t ) d t$.
(a) Find the values of $g ( 2 )$ and $g ( - 2 )$.
(b) For each of $g ^ { \prime } ( - 3 )$ and $g ^ { \prime \prime } ( - 3 )$, find the value or state that it does not exist.
(c) Find the $x$-coordinate of each point at which the graph of $g$ has a horizontal tangent line. For each of these points, determine whether $g$ has a relative minimum, relative maximum, or neither a minimum nor a maximum at the point. Justify your answers.
(d) For $- 4 < x < 3$, find all values of $x$ for which the graph of $g$ has a point of inflection. Explain your reasoning.
ap-calculus-bc 2014 Q3 View
The function $f$ is defined on the closed interval $[ - 5, 4 ]$. The graph of $f$ consists of three line segments and is shown in the figure above. Let $g$ be the function defined by $g ( x ) = \int _ { - 3 } ^ { x } f ( t ) \, dt$.
(a) Find $g ( 3 )$.
(b) On what open intervals contained in $- 5 < x < 4$ is the graph of $g$ both increasing and concave down? Give a reason for your answer.
(c) The function $h$ is defined by $h ( x ) = \frac { g ( x ) } { 5 x }$. Find $h ^ { \prime } ( 3 )$.
(d) The function $p$ is defined by $p ( x ) = f \left( x ^ { 2 } - x \right)$. Find the slope of the line tangent to the graph of $p$ at the point where $x = - 1$.
ap-calculus-bc 2016 Q3 View
The figure above shows the graph of the piecewise-linear function $f$. For $- 4 \leq x \leq 12$, the function $g$ is defined by $g ( x ) = \int _ { 2 } ^ { x } f ( t ) \, d t$.
(a) Does $g$ have a relative minimum, a relative maximum, or neither at $x = 10$ ? Justify your answer.
(b) Does the graph of $g$ have a point of inflection at $x = 4$ ? Justify your answer.
(c) Find the absolute minimum value and the absolute maximum value of $g$ on the interval $- 4 \leq x \leq 12$. Justify your answers.
(d) For $- 4 \leq x \leq 12$, find all intervals for which $g ( x ) \leq 0$.
ap-calculus-bc 2019 Q3 View
The continuous function $f$ is defined on the closed interval $- 6 \leq x \leq 5$. The figure above shows a portion of the graph of $f$, consisting of two line segments and a quarter of a circle centered at the point $( 5, 3 )$. It is known that the point $( 3, 3 - \sqrt { 5 } )$ is on the graph of $f$.
(a) If $\int _ { - 6 } ^ { 5 } f ( x ) \, dx = 7$, find the value of $\int _ { - 6 } ^ { - 2 } f ( x ) \, dx$. Show the work that leads to your answer.
(b) Evaluate $\int _ { 3 } ^ { 5 } \left( 2 f ^ { \prime } ( x ) + 4 \right) dx$.
(c) The function $g$ is given by $g ( x ) = \int _ { - 2 } ^ { x } f ( t ) \, dt$. Find the absolute maximum value of $g$ on the interval $- 2 \leq x \leq 5$. Justify your answer.
(d) Find $\lim _ { x \rightarrow 1 } \frac { 10 ^ { x } - 3 f ^ { \prime } ( x ) } { f ( x ) - \arctan x }$.
ap-calculus-bc 2024 Q4 View
The graph of the differentiable function $f$, shown for $-6 \leq x \leq 7$, has a horizontal tangent at $x = -2$ and is linear for $0 \leq x \leq 7$. Let $R$ be the region in the second quadrant bounded by the graph of $f$, the vertical line $x = -6$, and the $x$- and $y$-axes. Region $R$ has area 12.
(a) The function $g$ is defined by $g(x) = \int_{0}^{x} f(t)\, dt$. Find the values of $g(-6)$, $g(4)$, and $g(6)$.
(b) For the function $g$ defined in part (a), find all values of $x$ in the interval $0 \leq x \leq 6$ at which the graph of $g$ has a critical point. Give a reason for your answer.
(c) The function $h$ is defined by $h(x) = \int_{-6}^{x} f'(t)\, dt$. Find the values of $h(6)$, $h'(6)$, and $h''(6)$. Show the work that leads to your answers.
ap-calculus-bc 2025 Q4 View
The continuous function $f$ is defined on the closed interval $- 6 \leq x \leq 12$. The graph of $f$, consisting of two semicircles and one line segment, is shown in the figure.
Let $g$ be the function defined by $g ( x ) = \int _ { 6 } ^ { x } f ( t ) d t$.
A. Find $g ^ { \prime } ( 8 )$. Give a reason for your answer.
B. Find all values of $x$ in the open interval $- 6 < x < 12$ at which the graph of $g$ has a point of inflection. Give a reason for your answer.
C. Find $g ( 12 )$ and $g ( 0 )$. Label your answers.
D. Find the value of $x$ at which $g$ attains an absolute minimum on the closed interval $- 6 \leq x \leq 12$. Justify your answer.
bac-s-maths 2019 Q2 6 marks View
Below is the graphical representation $\mathscr { C } _ { g }$ in an orthogonal coordinate system of a function $g$ defined and continuous on $\mathbb { R }$. The curve $\mathscr { C } _ { g }$ is symmetric with respect to the $y$-axis and lies in the half-plane $y > 0$.
For all $t \in \mathbb { R }$ we define: $$G ( t ) = \int _ { 0 } ^ { t } g ( u ) \mathrm { d } u$$
Part A
The justifications of the answers to the following questions may be based on graphical considerations.
  1. Is the function $G$ increasing on $[ 0 ; + \infty [$ ? Justify.
  2. Justify graphically the inequality $G ( 1 ) \leqslant 0.9$.
  3. Is the function $G$ positive on $\mathbb { R }$ ? Justify.

In the rest of the problem, the function $g$ is defined on $\mathbb { R }$ by $g ( u ) = \mathrm { e } ^ { - u ^ { 2 } }$.
Part B
  1. Study of $g$ a. Determine the limits of the function $g$ at the boundaries of its domain. b. Calculate the derivative of $g$ and deduce the table of variations of $g$ on $\mathbb { R }$. c. Specify the maximum of $g$ on $\mathbb { R }$. Deduce that $g ( 1 ) \leqslant 1$.
  2. We denote $E$ the set of points $M$ located between the curve $\mathscr { C } _ { g }$, the $x$-axis and the lines with equations $x = 0$ and $x = 1$. We call $I$ the area of this set. We recall that: $$I = G ( 1 ) = \int _ { 0 } ^ { 1 } g ( u ) \mathrm { d } u$$ We wish to estimate the area $I$ by the method called ``Monte-Carlo'' described below.
    • We choose a point $M ( x ; y )$ by randomly drawing its coordinates $x$ and $y$ independently according to the uniform distribution on the interval $[ 0 ; 1 ]$. It is admitted that the probability that the point $M$ belongs to the set $E$ is equal to $I$.
    • We repeat $n$ times the experiment of choosing a point $M$ at random. We count the number $c$ of points belonging to the set $E$ among the $n$ points obtained.
    • The frequency $f = \frac { c } { n }$ is an estimate of the value of $I$. a. The figure below illustrates the method presented for $n = 100$. Determine the value of $f$ corresponding to this graph. b. The execution of the algorithm below uses the Monte-Carlo method described previously to determine a value of the number $f$. Copy and complete this algorithm. $f , x$ and $y$ are real numbers, $n , c$ and $i$ are natural integers. ALEA is a function that randomly generates a number between 0 and 1. \begin{verbatim} $c \leftarrow 0$ For $i$ varying from 1 to $n$ do: $x \leftarrow$ ALEA $y \leftarrow$ ALEA If $y \leqslant \ldots$ then $c \leftarrow \ldots$ end If end For $f \leftarrow \ldots$ \end{verbatim} c. An execution of the algorithm for $n = 1000$ gives $f = 0.757$. Deduce a confidence interval, at the 95\% confidence level, for the exact value of $I$.

Part C
We recall that the function $g$ is defined on $\mathbb { R }$ by $g ( u ) = \mathrm { e } ^ { - u ^ { 2 } }$ and that the function $G$ is defined on $\mathbb { R }$ by: $$G ( t ) = \int _ { 0 } ^ { t } g ( u ) \mathrm { d } u$$ We propose to determine an upper bound for $G ( t )$ for $t \geqslant 1$.
  1. A preliminary result. It is admitted that, for all real $u \geqslant 1$, we have $g ( u ) \leqslant \frac { 1 } { u ^ { 2 } }$. Deduce that, for all real $t \geqslant 1$, we have: $$\int _ { 1 } ^ { t } g ( u ) \mathrm { d } u \leqslant 1 - \frac { 1 } { t }$$
  2. Show that, for all real $t \geqslant 1$, $$G ( t ) \leqslant 2 - \frac { 1 } { t }$$ What can we say about the possible limit of $G ( t )$ as $t$ tends to $+ \infty$ ?
cmi-entrance 2024 QB2 15 marks View
(a) Find the domain of the function $g(x)$ defined by the following formula. $$g(x) = \int_{10}^{x} \log_{10}\left(\log_{10}\left(t^2 - 1000t + 10^{1000}\right)\right) dt$$ Calculate the quantities below. You may give an approximate answer where necessary, but clearly state which answers are exact and which are approximations.
(b) $g(1000)$.
(c) $x$ in $[10, 1000]$ where $g(x)$ has the maximum possible slope.
(d) $x$ in $[10, 1000]$ where $g(x)$ has the least possible slope.
(e) $\lim_{x \rightarrow \infty} \frac{\ln(x)}{g(x)}$ if it exists.
csat-suneung 2013 Q21 4 marks View
For the cubic function $f(x) = x^3 - 3x + a$, the function $$F(x) = \int_{0}^{x} f(t)\, dt$$ has exactly one extremum. What is the minimum value of the positive number $a$? [4 points]
(1) 1
(2) 2
(3) 3
(4) 4
(5) 5
csat-suneung 2026 Q15 4 marks View
The function $f ( x )$ is $$f ( x ) = \begin{cases} - x ^ { 2 } & ( x < 0 ) \\ x ^ { 2 } - x & ( x \geq 0 ) \end{cases}$$ and for a positive number $a$, the function $g ( x )$ is $$g ( x ) = \left\{ \begin{array} { c l } a x + a & ( x < - 1 ) \\ 0 & ( - 1 \leq x < 1 ) \\ a x - a & ( x \geq 1 ) \end{array} \right.$$ Let $k$ be the maximum value of $a$ such that the function $h ( x ) = \int _ { 0 } ^ { x } ( g ( t ) - f ( t ) ) d t$ has exactly one extremum. When $a = k$, what is the value of $k + h ( 3 )$? [4 points]
(1) $\frac { 9 } { 2 }$
(2) $\frac { 11 } { 2 }$
(3) $\frac { 13 } { 2 }$
(4) $\frac { 15 } { 2 }$
(5) $\frac { 17 } { 2 }$
isi-entrance 2013 Q36 4 marks View
Which of the following graphs represents the function $$f ( x ) = \int _ { 0 } ^ { \sqrt { x } } e ^ { - u ^ { 2 } / x } d u , \quad \text { for } \quad x > 0 \quad \text { and } \quad f ( 0 ) = 0 ?$$ (A), (B), (C), (D) as given by the respective graphs.
jee-advanced 2006 Q20 View
20. $\mathrm { f } ( \mathrm { x } ) = \left\{ \begin{array} { l l } \mathrm { e } ^ { \mathrm { x } } , & 0 \leq \mathrm { x } \leq 1 \\ 2 - \mathrm { e } ^ { \mathrm { x } - 1 } , & 1 < \mathrm { x } \leq 2 \\ \mathrm { x } - \mathrm { e } , & 2 < \mathrm { x } \leq 3 \end{array} \right.$ and $\mathrm { g } ( \mathrm { x } ) = \int _ { 0 } ^ { \mathrm { x } } \mathrm { f } ( \mathrm { t } ) \mathrm { dt } , \mathrm { x } \in [ 1,3 ]$ then $\mathrm { g } ( \mathrm { x } )$ has
(A) local maxima at $\mathrm { x } = 1 + \ln 2$ and local minima at $\mathrm { x } = \mathrm { e }$
(B) local maxima at $\mathrm { x } = 1$ and local minima at $\mathrm { x } = 2$
(C) no local maxima
(D) no local minima
Sol. (A), (B) $\mathrm { g } ^ { \prime } ( \mathrm { x } ) = \mathrm { f } ( \mathrm { x } ) = \begin{cases} \mathrm { e } ^ { \mathrm { x } } & 0 \leq \mathrm { x } \leq 1 \\ 2 - \mathrm { e } ^ { \mathrm { x } - 1 } & 1 < \mathrm { x } \leq 2 \\ \mathrm { x } - \mathrm { e } & 2 < \mathrm { x } \leq 3 \end{cases}$ $\mathrm { g } ^ { \prime } ( \mathrm { x } ) = 0$, when $\mathrm { x } = 1 + \ln 2$ and $\mathrm { x } = \mathrm { e }$ $g ^ { \prime \prime } ( x ) = \left\{ \begin{array} { c c } - e ^ { x - 1 } & 1 < x \leq 2 \\ 1 & 2 < x \leq 3 \end{array} \right\}$ $\mathrm { g } ^ { \prime \prime } ( 1 + \ln 2 ) = - \mathrm { e } ^ { \ln 2 } < 0$ hence at $\mathrm { x } = 1 + \ln 2 , \mathrm {~g} ( \mathrm { x } )$ has a local maximum $\mathrm { g } ^ { \prime \prime } ( \mathrm { e } ) = 1 > 0$ hence at $\mathrm { x } = \mathrm { e } , \mathrm { g } ( \mathrm { x } )$ has local minimum. ∵ $\mathrm { f } ( \mathrm { x } )$ is discontinuous at $\mathrm { x } = 1$, then we get local maxima at $\mathrm { x } = 1$ and local minima at $\mathrm { x } = 2$.
Section - C
Comprehension I
There are $n$ urns each containing $n + 1$ balls such that the $i$ th urn contains $i$ white balls and ( $n + 1 - i$ ) red balls. Let $u _ { i }$ be the event of selecting ith urn, $\mathrm { i } = 1,2,3 \ldots , \mathrm { n }$ and w denotes the event of getting a white ball.
jee-advanced 2008 Q16 View
Consider the function $f : ( - \infty , \infty ) \rightarrow ( - \infty , \infty )$ defined by
$$f ( x ) = \frac { x ^ { 2 } - a x + 1 } { x ^ { 2 } + a x + 1 } , 0 < a < 2 .$$
Let
$$g ( x ) = \int _ { 0 } ^ { e ^ { x } } \frac { f ^ { \prime } ( t ) } { 1 + t ^ { 2 } } d t$$
Which of the following is true?
(A) $g ^ { \prime } ( x )$ is positive on $( - \infty , 0 )$ and negative on $( 0 , \infty )$
(B) $g ^ { \prime } ( x )$ is negative on $( - \infty , 0 )$ and positive on $( 0 , \infty )$
(C) $g ^ { \prime } ( x )$ changes sign on both $( - \infty , 0 )$ and $( 0 , \infty )$
(D) $g ^ { \prime } ( x )$ does not change sign on $( - \infty , \infty )$
jee-advanced 2010 Q34 View
The value of $\lim _ { x \rightarrow 0 } \frac { 1 } { x ^ { 3 } } \int _ { 0 } ^ { x } \frac { t \ln ( 1 + t ) } { t ^ { 4 } + 4 } d t$ is
A) 0
B) $\frac { 1 } { 12 }$
C) $\frac { 1 } { 24 }$
D) $\frac { 1 } { 64 }$
jee-advanced 2024 Q13 4 marks View
Let the function $f : [ 1 , \infty ) \rightarrow \mathbb { R }$ be defined by $$f ( t ) = \left\{ \begin{array} { c c } ( - 1 ) ^ { n + 1 } 2 , & \text { if } t = 2 n - 1 , n \in \mathbb { N } , \\ \frac { ( 2 n + 1 - t ) } { 2 } f ( 2 n - 1 ) + \frac { ( t - ( 2 n - 1 ) ) } { 2 } f ( 2 n + 1 ) , & \text { if } 2 n - 1 < t < 2 n + 1 , n \in \mathbb { N } . \end{array} \right.$$ Define $g ( x ) = \int _ { 1 } ^ { x } f ( t ) d t , x \in ( 1 , \infty )$. Let $\alpha$ denote the number of solutions of the equation $g ( x ) = 0$ in the interval $( 1,8 ]$ and $\beta = \lim _ { x \rightarrow 1 + } \frac { g ( x ) } { x - 1 }$. Then the value of $\alpha + \beta$ is equal to $\_\_\_\_$ .
jee-main 2017 Q68 View
If $f : \mathbb { R } \to \mathbb { R }$ is a differentiable function and $f ( 2 ) = 6$, then $\lim _ { x \to 2 } \int _ { 6 } ^ { f ( x ) } \frac { 2 t \, d t } { ( x - 2 ) }$ is:
(1) $2 f ^ { \prime } ( 2 )$
(2) $12 f ^ { \prime } ( 2 )$
(3) $0$
(4) $24 f ^ { \prime } ( 2 )$
jee-main 2024 Q67 View
$\lim _ { x \rightarrow \frac { \pi } { 2 } } \left( \frac { 1 } { \left( x - \frac { \pi } { 2 } \right) ^ { 2 } } \int _ { x ^ { 3 } } ^ { \left( \frac { \pi } { 2 } \right) ^ { 3 } } \cos \left( \frac { 1 } { t ^ { 3 } } \right) d t \right)$ is equal to
(1) $\frac { 3 \pi } { 8 }$
(2) $\frac { 3 \pi ^ { 2 } } { 4 }$
(3) $\frac { 3 \pi ^ { 2 } } { 8 }$
(4) $\frac { 3 \pi } { 4 }$
jee-main 2024 Q68 View
Let $f ( x ) = \int _ { 0 } ^ { x } \left( t + \sin \left( 1 - e ^ { t } \right) \right) d t , x \in \mathbb { R }$. Then, $\lim _ { x \rightarrow 0 } \frac { f ( x ) } { x ^ { 3 } }$ is equal to
(1) $- \frac { 1 } { 6 }$
(2) $\frac { 2 } { 3 }$
(3) $- \frac { 2 } { 3 }$
(4) $\frac { 1 } { 6 }$