Compute a definite integral of a piecewise-defined or periodic function, requiring splitting the domain, using periodicity, or handling different cases across subintervals.
The function $f$ is defined by $f ( x ) = \left\{ \begin{array} { l l } 2 & \text { for } x < 3 \\ x - 1 & \text { for } x \geq 3 \end{array} \right.$. What is the value of $\int _ { 1 } ^ { 5 } f ( x ) d x$ ? (A) 2 (B) 6 (C) 8 (D) 10 (E) 12
The graph of the piecewise linear function $f$ is shown above. What is the value of $\int _ { - 1 } ^ { 9 } ( 3 f ( x ) + 2 ) d x$ ? (A) 7.5 (B) 9.5 (C) 27.5 (D) 47 (E) 48.5
Calculate the following integrals whenever possible. If a given integral does not exist, state so. Note that $[ x ]$ denotes the integer part of $x$, i.e., the unique integer $n$ such that $n \leq x < n + 1$. a) $\int _ { 1 } ^ { 4 } x ^ { 2 } d x$ Answer: $\_\_\_\_$ b) $\int _ { 1 } ^ { 3 } [ x ] ^ { 2 } d x$ Answer: $\_\_\_\_$ c) $\int _ { 1 } ^ { 2 } \left[ x ^ { 2 } \right] d x$ Answer: $\_\_\_\_$ d) $\int _ { - 1 } ^ { 1 } \frac { 1 } { x ^ { 2 } } d x$ Answer: $\_\_\_\_$
The function $f ( x )$ satisfies $f ( x + 3 ) = f ( x )$ for all real numbers $x$, and $$f ( x ) = \begin{cases} x & ( 0 \leq x < 1 ) \\ 1 & ( 1 \leq x < 2 ) \\ - x + 3 & ( 2 \leq x < 3 ) \end{cases}$$ If $\int _ { - a } ^ { a } f ( x ) d x = 13$, what is the value of the constant $a$? [4 points] (1) 10 (2) 12 (3) 14 (4) 16 (5) 18
For the quadratic function $f ( x ) = \frac { 3 x - x ^ { 2 } } { 2 }$, a function $g ( x )$ defined on the interval $[ 0 , \infty )$ satisfies the following conditions. (가) When $0 \leq x < 1$, $g ( x ) = f ( x )$. (나) When $n \leq x < n + 1$, $$\begin{aligned}
& g ( x ) = \frac { 1 } { 2 ^ { n } } \{ f ( x - n ) - ( x - n ) \} + x \\
& \text{(Here, } n \text{ is a natural number.)}
\end{aligned}$$ For some natural number $k ( k \geq 6 )$, the function $h ( x )$ is defined as $$h ( x ) = \begin{cases} g ( x ) & ( 0 \leq x < 5 \text{ or } x \geq k ) \\ 2 x - g ( x ) & ( 5 \leq x < k ) \end{cases}$$ When the sequence $\left\{ a _ { n } \right\}$ is defined by $a _ { n } = \int _ { 0 } ^ { n } h ( x ) d x$ and $\lim _ { n \rightarrow \infty } \left( 2 a _ { n } - n ^ { 2 } \right) = \frac { 241 } { 768 }$, find the value of $k$. [4 points]
An increasing continuous function $f ( x )$ on the set of all real numbers satisfies the following conditions. (가) For all real numbers $x$, $f ( x ) = f ( x - 3 ) + 4$. (나) $\int _ { 0 } ^ { 6 } f ( x ) d x = 0$ What is the area enclosed by the graph of $y = f ( x )$, the $x$-axis, and the two lines $x = 6$ and $x = 9$? [4 points] (1) 9 (2) 12 (3) 15 (4) 18 (5) 21
For the function $f ( x ) = \pi \sin 2 \pi x$, a function $g ( x )$ with domain being the set of all real numbers and range being the set $\{ 0,1 \}$, and a natural number $n$ satisfy the following conditions. What is the value of $n$? [4 points] The function $h ( x ) = f ( n x ) g ( x )$ is continuous on the set of all real numbers and $$\int _ { - 1 } ^ { 1 } h ( x ) d x = 2 , \quad \int _ { - 1 } ^ { 1 } x h ( x ) d x = - \frac { 1 } { 32 }$$ (1) 8 (2) 10 (3) 12 (4) 14 (5) 16
A function $f ( x )$ that is continuous on the set of all real numbers satisfies the following condition. When $n - 1 \leq x < n$, $| f ( x ) | = | 6 ( x - n + 1 ) ( x - n ) |$. (Here, $n$ is a natural number.) For the function $$g ( x ) = \int _ { 0 } ^ { x } f ( t ) d t - \int _ { x } ^ { 4 } f ( t ) d t$$ defined on the open interval $(0, 4)$, when $g ( x )$ has a minimum value of 0 at $x = 2$, what is the value of $\int _ { \frac { 1 } { 2 } } ^ { 4 } f ( x ) d x$? [4 points] (1) $- \frac { 3 } { 2 }$ (2) $- \frac { 1 } { 2 }$ (3) $\frac { 1 } { 2 }$ (4) $\frac { 3 } { 2 }$ (5) $\frac { 5 } { 2 }$
Let $g : [ 0,1 ] \rightarrow \mathbb { R }$ be the function such that $g ( x ) = 1 / x$ if $x \geqslant \mathrm { e } ^ { - 1 }$ and $g ( x ) = 0$ otherwise. We fix a real $\varepsilon \in ] 0 , \mathrm { e } ^ { - 1 } [$. We define two continuous applications $g ^ { + } , g ^ { - } : [ 0,1 ] \rightarrow \mathbb { R }$ as follows:
$g ^ { + }$ is affine on $\left[ \mathrm { e } ^ { - 1 } - \varepsilon , \mathrm { e } ^ { - 1 } \right]$ and coincides with $g$ on $\left[ 0 , \mathrm { e } ^ { - 1 } - \varepsilon \right] \cup \left[ \mathrm { e } ^ { - 1 } , 1 \right]$;
$g ^ { - }$ is affine on $\left[ \mathrm { e } ^ { - 1 } , \mathrm { e } ^ { - 1 } + \varepsilon \right]$ and coincides with $g$ on $\left[ 0 , \mathrm { e } ^ { - 1 } \left[ \cup \left[ \mathrm { e } ^ { - 1 } + \varepsilon , 1 \right] \right. \right.$.
Calculate $\int _ { 0 } ^ { 1 } g ^ { + } ( t ) \mathrm { d } t$ and $\int _ { 0 } ^ { 1 } g ^ { - } ( t ) \mathrm { d } t$.
Let $h$ be a function from $\mathbb{R}$ to $\mathbb{R}$, continuous and $2\pi$-periodic on $\mathbb{R}$. For any complex number $z$ such that $|z| < 1$, $$g(z) = \frac{1}{2\pi} \int_0^{2\pi} h(t) \mathcal{P}(t,z) \, \mathrm{d}t \quad \text{where} \quad \mathcal{P}(t,z) = \operatorname{Re}\left(\frac{\mathrm{e}^{\mathrm{i}t} + z}{\mathrm{e}^{\mathrm{i}t} - z}\right)$$ Let $\varphi \in \mathbb{R}$. Show that, for any complex number $z$ such that $|z| < 1$, $g(z) = \frac{1}{2\pi} \int_{\varphi}^{\varphi + 2\pi} h(t) \mathcal{P}(t,z) \, \mathrm{d}t$.
The function $q$ associates to any real $x$ the real number $q ( x ) = x - \lfloor x \rfloor - \frac { 1 } { 2 }$, where $\lfloor x \rfloor$ denotes the integer part of $x$. Show that for all integer $n > 1$, $$\int _ { 1 } ^ { n } \frac { q ( u ) } { u } \mathrm {~d} u = \ln ( n ! ) + ( n - 1 ) - n \ln ( n ) - \frac { 1 } { 2 } \ln ( n ) = \ln \left( \frac { n ! e ^ { n } } { n ^ { n } \sqrt { n } } \right) - 1$$
Let $[ x ]$ denote the largest integer less than or equal to $x$. Then $\int_0^{n^{1/k}} \left[ x ^ { k } + n \right] dx$ equals (a) $n ^ { 2 } + \sum_{i=1}^{n} i ^ { 1 / k }$ (b) $2 n ^ { ( 1 + k ) / k } - \sum_{i=1}^{n} i ^ { 1 / k }$ (c) $2 n ^ { ( 1 + k ) / k } - \sum_{i=1}^{n-1} i ^ { 1 / k }$ (d) None of these.
6. Let $f ( x ) = x - [ x ]$, for every real number $x$, where $[ x ]$ is the integral part of $x$. Then $\int 1 - 1 f ( x ) d x$ is : (A) 1 (B) 2 (C) 0 (D) $1 / 2$
8. If for a real number $y , [ y ]$ is the greatest integer less than or equal to $y$, then the value of the integral $\int \sqcap / 23 \sqcap / 2 [ 2 \sin x . d x ]$ is: (A) $- \pi$ (B) 0 (C) $- \pi / 2$ (D) $\pi / 2$