Antiderivatives and Indefinite Integrals
Definition of an Antiderivative
A function $F$ is called an antiderivative of $f$ on an interval $I$ if $F^{\prime}(x)=f(x)$ for all $x$ on $I$.
Representation Antiderivatives
If $F$ is an antiderivative of $f$ on an interval $I$, then $F(x)+C$ represents the most general antiderivative of $f$ on $I$, where $C$ is a constant.
Example 1
- Find an antiderivative for each of the following functions.
- $f(x)=3 x^2$
- $g(x)=\cos x+3$
▶️Answer/Explanation
Solution
a. $F(x)=x^3+C$ Derivative of $x^3$ is $3 x^2$.
b. $G(x)=\sin x+3 x+C$ Derivative of $\sin x$ is $\cos x$, and derivative of $3 x$ is 3 .
Definition of Indefinite Integral
The set of all antiderivatives of $f$ is the indefinite integral of $f$ with respect to $x$ denoted by
$
\int f(x) d x
$
Thus $\int f(x) d x=F(x)+C$ means $F^{\prime}(x)=f(x)$.
Example 2
- Find the general solution of $F^{\prime}(x)=\sec ^2 x$.
▶️Answer/Explanation
Solution
$F(x)=\int \sec ^2 x=\tan x+C \quad$ Derivative of $\tan x$ is $\sec ^2 x$.
Table of Indefinite Integrals
- $\int k d x=k x+C \quad \quad \int k f(x) d x=k \int f(x) d x $
- $\int[f(x) \pm g(x)] d x=\int f(x) d x \pm \int g(x) d x $
- $\int x^n d x=\frac{x^{n+1}}{n+1}+C, n \neq-1 \quad \text { Power Rule } $
- $\int e^x d x=e^x+C $
- $\int \sin x d x=-\cos x+C \quad \quad \int \cos x d x=\sin x+C $
- $\int \sec ^2 x d x=\tan x+C \quad \quad \int \csc ^2 x d x=-\cot x+C $
- $\int \sec x \tan x d x=\sec x+C \quad \quad \int \csc ^2 x \cot x d x=-\csc x+C$
Example 3
- Find the antiderivative of $x^3-3 x+2$.
▶️Answer/Explanation
Solution
$
\begin{aligned}
& \int\left(x^3-3 x+2\right) d x=\int x^3 d x-\int 3 x d x+\int 2 d x \\
& =\frac{x^{3+1}}{3+1}-3 \cdot \frac{x^{1+1}}{1+1}+2 x+C \quad \text { Power Rule } \\
& =\frac{x^4}{4}-\frac{3 x^2}{2}+2 x+C \quad \text { Answer } \\
& \text { Check: } \frac{d}{d x}\left(\frac{x^4}{4}-\frac{3 x^2}{2}+2 x+C\right)=\frac{1}{4} \cdot 4 x^3-\frac{3}{2} \cdot 2 x+2 \cdot 1+0 \\
& =x^3-3 x+2 \quad \checkmark
\end{aligned}
$
Example 4
- Find the general indefinite integral
$
\int(\sqrt{x}-\sec x \tan x) d x \text {. }
$
▶️Answer/Explanation
Solution
$
\begin{aligned}
& \int(\sqrt{x}-\sec x \tan x) d x=\int \sqrt{x} d x-\int \sec x \tan x d x \\
& =\frac{x^{\frac{1}{2}+1}}{\frac{1}{2}+1}-\sec x+C=\frac{2}{3} x^{\frac{3}{2}}-\sec x+C
\end{aligned}
$
$
\text { Check: } \begin{aligned}
& \frac{d}{d x}\left(\frac{2}{3} x^{\frac{3}{2}}-\sec x+C\right)=\frac{2}{3} \cdot \frac{3}{2} x^{\frac{3}{2}-1}-\sec x \tan x+0 \\
= & x^{\frac{1}{2}}-\sec x \tan x=\sqrt{x}-\sec x \tan x \quad \sqrt{ }
\end{aligned}
$
Example5
- If $\frac{d y}{d x}=3 x^2-1$, and if $y=-1$ when $x=1$, then $y=$
(A) $x^3-x+1$
(B) $x^3-x-1$
(C) $-x^3+x-1$
(D) $-x^3+1$
▶️Answer/Explanation
Ans:B
- Which of the following is the antiderivative of $f(x)=\tan x$ ?
(A) $\sec x+\tan x+C$
(B) $\csc x+\cot x+C$
(C) $\ln |\csc x|+C$
(D) $-\ln |\cos x|+C$
▶️Answer/Explanation
Ans:D
- A curve has a slope of $-x+2$ at each point $(x, y)$ on the curve. Which of the following is an equation for this curve if it passes through the point $(2,1)$ ?
(A) $\frac{1}{2} x^2-2 x-4$
(B) $2 x^2+x-8$
(C) $-\frac{1}{2} x^2+2 x-1$
(D) $x^2-2 x+1$
▶️Answer/Explanation
Ans:C
Riemann Sum and Area Approximation
The area of a region $S$ that lies under the curve $y=f(x)$ from $a$ to $b$ can be approximated by summing the areas of a collection of rectangles.
Figure 4.1 The rectangles approximate the area between the graph of the function $y=f(x)$ and the $x$-axis.
Definition of Riemann Sum
Let $f$ be a continuous function defined on the closed interval $[a, b]$, and let $\Delta$ be a partition of $[a, b]$ given by
$
a=x_0<x_1<x_2<\cdots<x_{n-1}<x_n=b
$
where $\Delta x_i$ is the width of the $i$ th interval. If $c_i$ is any point in the $i$ th interval, then the sum
$
\sum_{i=1}^n f\left(c_i\right) \Delta x_i=f\left(c_1\right) \Delta x_1+f\left(c_2\right) \Delta x_2+\cdots+f\left(c_i\right) \Delta x_i+\cdots+f\left(c_n\right) \Delta x_n
$
is called a Riemann sum for $f$ on the interval $[a, b]$.
If every subinterval is of equal width, the partition is regular and $\Delta x=\frac{\boldsymbol{b}-\boldsymbol{a}}{\boldsymbol{n}}$.
Then the Riemann sum can be written
$
\sum_{i=1}^n f\left(c_i\right) \Delta x=\Delta x\left[f\left(c_1\right)+f\left(c_2\right)+\cdots+f\left(c_i\right)+\cdots+f\left(c_n\right)\right]
$
where $c_i=a+i(\Delta x)$.
Left, Right, and Midpoint Riemann Sum Approximation
If $c_i$ is the left endpoint of each subinterval then $\sum_{i=1}^n f\left(c_i\right) \Delta x_i$ is called a left Riemann sum.
If $c_i$ is the right endpoint of each subinterval then $\sum_{i=1}^n f\left(c_i\right) \Delta x_i$ is called a right Riemann sum.
If $c_i$ is the midpoint of each subinterval then $\sum_{i=1}^n f\left(c_i\right) \Delta x_i$ is called a midpoint Riemann sum.
Example1
- Approximate the area of the region bounded by the graph of $f(x)=-x^2+x+2$, the $x$-axis, and the vertical lines $x=0$ and $x=2$,
(a) by using a left Riemann sum with four subintervals,
(b) by using a right Riemann sum with four subintervals, and
(c) by using a midpoint Riemann sum with four subintervals.
▶️Answer/Explanation
Solution
(a)
For a left Riemann sum with 4 subintervals, we use the four rectangles whose heights are the values of $f$ at the left endpoints of their bases.
The left endpoints of each subinterval are $0,0.5,1$, and 1.5 and $\Delta x=\frac{b-a}{n}=\frac{2-0}{4}=\frac{1}{2}$.
The left Riemann sum is
$
\begin{aligned}
\sum_{i=1}^4 f\left(c_i\right) \Delta x_i & =f(0) \cdot\left(\frac{1}{2}\right)+f(0.5) \cdot\left(\frac{1}{2}\right)+f(1) \cdot\left(\frac{1}{2}\right)+f(1.5) \cdot\left(\frac{1}{2}\right) \\
& =\frac{1}{2}[2+2.25+2+1.25]=3.75
\end{aligned}
$
(b) For a right Riemann sum with 4 subintervals, we use the three rectangles whose heights are the values of $f$ at the right endpoints of their bases.
The right endpoints of each subinterval are $0.5,1,1.5$, and 2 , and $\Delta x=\frac{b-a}{n}=\frac{2-0}{4}=\frac{1}{2}$.
The right Riemann sum is
$
\begin{aligned}
\sum_{i=1}^4 f\left(c_i\right) \Delta x_i & =f(0.5) \cdot\left(\frac{1}{2}\right)+f(1) \cdot\left(\frac{1}{2}\right)+f(1.5) \cdot\left(\frac{1}{2}\right)+f(2) \cdot\left(\frac{1}{2}\right) \\
& =\frac{1}{2}[2.25+2+1.25+0]=2.75
\end{aligned}
$
(c) For a midpoint Riemann sum with 4 subintervals, we use the four rectangles whose heights are the values of $f$ at the midpoints of their bases.
The midpoints of each subinterval are $0.25,0.75,1.25$, and 1.75 , and $\Delta x=\frac{b-a}{n}=\frac{2-0}{4}=\frac{1}{2}$. The midpoint Riemann sum is
$
\begin{aligned}
\sum_{i=1}^4 f\left(c_i\right) \Delta x_i & =f(.25) \cdot\left(\frac{1}{2}\right)+f(.75) \cdot\left(\frac{1}{2}\right)+f(1.25) \cdot\left(\frac{1}{2}\right)+f(1.75) \cdot\left(\frac{1}{2}\right) \\
& =\frac{1}{2}[2.1875+2.1875+1.6875+0.6875]=3.375
\end{aligned}
$
Example 2
The function $f$ is continuous on the closed interval $[0,12]$ and has values as shown in the table above. Use a midpoint Riemann sum with 4 subintervals of equal length to approximate the area that lies under $f$ and above the $x$-axis from $x=0$ to $x=12$.
▶️Answer/Explanation
Solution
The four intervals are $[0,3],[3,6],[6,9]$, and $[9,12]$.
$1.5,4.5,7.5$, and 10.5 are the midpoints of each interval.
Midpoint Riemann sum is
$
\begin{aligned}
& f(1.5) \cdot 3+f(4.5) \cdot 3+f(7.5) \cdot 3+f(10.5) \cdot 3 \\
& =3 \cdot[1.45+5.05+12.25+23.05] \\
& =125.4
\end{aligned}
$
Example3
- Using a left Riemann sum with three subintervals $[0,1],[1,2]$, and $[2,3]$, what is the approximation of $\int_0^3(3-x)(x+1) d x$ ?
(A) 7.5
(B) 9
(C) 10
(D) 11.5
▶️Answer/Explanation
Ans:C
Example4
The function $f$ is continuous on the closed interval $[1,10]$ and has values as shown in the table above. Using a right Riemann sum with four subintervals $[1,3],[3,5],[5,8],[8,10]$, what is the approximation of $\int_1^{10} f(x) d x$ ?
(A) 96
(B) 116
(C) 132
(D) 159
▶️Answer/Explanation
Ans:D
Example5
- The expression $\frac{1}{20}\left[\left(\frac{1}{20}\right)^2+\left(\frac{2}{20}\right)^2+\left(\frac{3}{20}\right)^2+\ldots+\left(\frac{20}{20}\right)^2\right]$ is a Riemann sum approximation for
(A) $\frac{1}{20} \int_0^{20} x^2 d x$
(B) $\frac{1}{20} \int_0^1 x^2 d x$
(C) $\int_0^1 x^2 d x$
(D) $\int_0^1 \frac{1}{x^2} d x$
▶️Answer/Explanation
Ans:C
Definite Integral, Area Under a Curve, and Application
Definition of a Definite Integral
If $f$ is a continuous function defined for $a \leq x \leq b$, then the definite integral of $f$ from $a$ to $b$ is
$
\int_a^b f(x) d x=\lim _{n \rightarrow \infty} \sum_{i=1}^n f\left(c_i\right) \Delta x,
$
where $c_i=a+i \Delta x$ and $\Delta x=(b-a) / n$.
If $y=f(x)$ is continuous and nonnegative over a closed interval $[a, b]$ then the area of the region bounded by the graph of $f$, the $x$-axis, and the vertical lines $x=a$ and $x=b$ is given by
$
\text { Area }=\int_a^b f(x) d x
$
If $f(x)$ takes on both positive and negative values over a closed interval $[a, b]$, then the area of the region bounded by the graph of $f$ and the $x$-axis is obtained by adding the absolute value of the definite integral over each subinterval where $f(x)$ does not change sign.
The figure above shows how both the total area and the value of definite integral can be interpreted in terms of areas between the graph of $f(x)$ and the $x$-axis.
The definite integral of $f(x)$ over $[a, b]=\int_a^b f(x) d x=A_1-A_2+A_3$.
The total area between the curve and the $x$-axis over $[a, b]=\int_a^b|f(x)| d x=A_1+A_2+A_3$.
Application of Definite Integral
If the population density of a region at a distance $x$ from a straight road is $D(x)$, then the total population of the region between $x=a$ and $x=b$ is given by
Total Population $=\int_a^b f(x) \cdot D(x) d x$
If the population density of a circular region at a distance $r$ from the center is $D(r)$, then the total population of the circular region between $r=a$ and $r=b$ is given by
Total Population $=2 \pi \int_a^b r D(r) d r$
Example1
- The figure on the right shows the graph of $f(x)=x^3-x^2-6 x$
(a) Find the definite integral of $f(x)$ on $[-2,3]$.
(b) Find the area between the graph of $f(x)$ and the $x$-axis on $[-2,3]$.
▶️Answer/Explanation
Solution
$
\text { (a) } \begin{aligned}
\int_{-2}^3\left(x^3-x^2-6 x\right) d x=\left[\frac{x^4}{4}-\frac{x^3}{3}-3 x^2\right]_{-2}^3 \\
=\left(\frac{81}{4}-\frac{27}{3}-27\right)-\left(\frac{16}{4}+\frac{8}{3}-12\right)=-\frac{125}{12}
\end{aligned}
$
$
\begin{aligned}
(b) & \text { Area }=\int_{-2}^3\left|x^3-x^2-6 x\right| d x=\left|\left[\frac{x^4}{4}-\frac{x^3}{3}-3 x^2\right]_{-2}^0\right|+\left|\left[\frac{x^4}{4}-\frac{x^3}{3}-3 x^2\right]_0^3\right| \\
& =\left|\frac{16}{3}\right|+\left|-\frac{63}{4}\right|=\frac{253}{12}
\end{aligned}
$
Example2
A rectangular region located beside a highway and between two straight roads 11 miles apart are shown in the figure above. The population density of the region at a distance $x$ miles from the highway is given by $D(x)=15 x \sqrt{x}-3 x^2$, where $0 \leq x \leq 25$. How many people live between 16 to 25 miles from the highway?
▶️Answer/Explanation
Population $=\int_a^b f(x) \cdot D(x) d x$
$=\int_{16}^{25} 11\left(15 x \sqrt{x}-3 x^2\right) d x \quad f(x)=11$ and $D(x)=15 x \sqrt{x}-3 x^2$
$\begin{aligned} & =\int_{16}^{25} 11\left(15 x^{3 / 2}-3 x^2\right) d x \\ & =11\left[15\left(\frac{2}{5}\right) x^{5 / 2}-x^3\right]_{16}^{25}=11\left[6 x^{5 / 2}-x^3\right]_{16}^{25} \\ & =11\left[\left\{6(25)^{5 / 2}-(25)^3\right\}-\left\{6(16)^{5 / 2}-(16)^3\right\}\right] \\ & =11[3,125-2,048] \\ & =11,847\end{aligned}$
Example3
- $
\int_0^3 \frac{d x}{\sqrt{1+x}}=
$
(A) 2
(B) 2.5
(C) 3
(D) 4
▶️Answer/Explanation
Ans:A
Example4
- The area of the region in the first quadrant enclosed by the graph of $f(x)=4 x-x^3$ and the $x$-axis is
(A) $\frac{11}{4}$
(B) $\frac{7}{2}$
(C) 4
(D) $\frac{11}{2}$
▶️Answer/Explanation
Ans:C
Example5
- $
\int_0^5 \sqrt{25-x^2} d x=
$
(A) $\frac{25 \pi}{8}$
(B) $\frac{25 \pi}{4}$
(C) $\frac{25 \pi}{2}$
(D) $25 \pi$
▶️Answer/Explanation
Ans:B
Example6
- The population density of a circular region is given by $f(r)=10-3 \sqrt{r}$ people per square mile, where $r$ is the distance from the center of the city, in miles. Which of the following expressions gives the number of people who live within a 3 mile radius from the center of the city?
(A) $\pi \int_0^3 r^2(10-3 \sqrt{r}) d r$
(B) $\pi \int_0^3(r+3)^2(10-3 \sqrt{r}) d r$
(C) $2 \pi \int_0^3(r+3)(10-3 \sqrt{r}) d r$
(D) $2 \pi \int_0^3 r(10-3 \sqrt{r}) d r$
▶️Answer/Explanation
Ans:D
Properties of Definite Integral
Definition
$
\int_a^a f(x) d x=0 \quad \int_a^b f(x) d x=-\int_b^a f(x) d x
$
Constant Multiple
$
\int_a^b c d x=c(b-a) \quad \int_a^b c f(x) d x=c \int_a^b f(x) d x
$
Sum and Difference
$
\int_a^b[f(x) \pm g(x)] d x=\int_a^b f(x) d x \pm \int_a^b g(x) d x
$
Additivity
$
\int_a^b f(x) d x+\int_b^c f(x) d x=\int_a^c f(x) d x
$
Integrals of Symmetric Functions
If $f$ is even $f(-x)=f(x)$, then $\int_{-a}^a f(x) d x=2 \int_0^a f(x) d x$
If $f$ is odd $f(-x)=-f(x)$, then $\int_{-a}^a f(x) d x=0$
Comparison Property
If $f(x) \geq 0$ for $a \leq x \leq b$, then $\int_a^b f(x) d x \geq 0$
If $f(x) \geq g(x)$ for $a \leq x \leq b$, then $\int_a^b f(x) d x \geq \int_a^b g(x) d x$
If $m \leq f(x) \leq M$ for $a \leq x \leq b$, then $m(b-a) \leq \int_a^b f(x) d x \leq M(b-a)$
Example1
- Suppose that $\int_{-3}^4 f(x) d x=5, \int_{-3}^4 g(x) d x=-4$, and $\int_{-3}^1 f(x) d x=2$.
Find (a) $\int_{-3}^4[2 f(x)-3 g(x)] d x$
(b) $\int_1^4 f(x) d x$
(c) $\int_{-3}^4[g(x)+2] d x$.
▶️Answer/Explanation
Solution
$
\begin{aligned}
& \int_{-3}^4[2 f(x)-3 g(x)] d x=\int_{-3}^4 2 f(x) d x-\int_{-3}^4 3 g(x) d x \\
& =2 \int_{-3}^4 f(x) d x-3 \int_{-3}^4 g(x) d x=2 \cdot 5-3 \cdot(-4)=22
\end{aligned}
$
(a)
$
\begin{aligned}
(b) & \int_{-3}^4 f(x) d x=\int_{-3}^1 f(x) d x+\int_1^4 f(x) d x \\
& \Rightarrow 5=2+\int_1^4 f(x) d x \Rightarrow \int_1^4 f(x) d x=3
\end{aligned}
$
(c) $\int_{-3}^4[g(x)+2] d x=\int_{-3}^4 g(x) d x+\int_{-3}^4 2 d x=-4+2(4+3)=10$
Example2
- If $\int_a^b f(x) d x=2 a-5 b$, then $\int_a^b[f(x)-2] d x=$
(A) $-7 b$
(B) $-3 b$
(C) $4 a-7 b$
(D) $4 a-3 b$
▶️Answer/Explanation
Ans:C
Example3
- If $\int_1^6 f(x) d x=\frac{15}{2}$ and $\int_6^4 f(x) d x=5$, then $\int_1^4 f(x) d x=$
(A) $\frac{5}{2}$
(B) $\frac{9}{2}$
(C) $\frac{19}{2}$
(D) $\frac{25}{2}$
▶️Answer/Explanation
Ans:C
Example4
- If $\int_{-2}^6 f(x) d x=10$ and $\int_2^6 f(x) d x=3$, then $\int_2^6 f(4-x) d x=$
(A) 3
(B) 6
(C) 7
(D) 10
▶️Answer/Explanation
Ans:B
Example5
The graph of $y=f(x)$ is shown in the figure above. If $A$ and $B$ are positive numbers that represent the areas of the shaded regions, what is the value of $\int_{-3}^3 f(x) d x-2 \int_{-1}^3 f(x) d x$, in terms of $A$ and $B$ ?
(A) $-A-B$
(B) $A+B$
(C) $A-2 B$
(D) $A-B$
▶️Answer/Explanation
Ans:D
Trapezoidal Rule
Let $f$ be continuous on $[a, b]$. The trapezoidal rule for approximating $\int_a^b f(x) d x$ is given by
$
\int_a^b f(x) d x=\frac{\Delta x}{2}\left[f\left(x_0\right)+2 f\left(x_1\right)+2 f\left(x_2\right)+\cdots+2 f\left(x_{n-1}\right)+f\left(x_n\right)\right],
$
where $x_0=a, x_1=a+\Delta x, x_2=a+2 \Delta x, \ldots, x_{n-1}=a+(n-1) \Delta x, x_n=b$, and $\Delta x=\frac{b-a}{n}$.
Example1
- Use the trapezoidal rule to approximate the integral $\int_1^3 \sqrt{1+x^2} d x$ with four subintervals.
▶️Answer/Explanation
Solution
$
\begin{aligned}
& \Delta x=\frac{b-a}{n}=\frac{3-1}{4}=\frac{1}{2} \\
& x_0=1, x_1=1+0.5=1.5, x_2=1+2(0.5)=2, x_3=1+3(0.5)=2.5, \\
& \text { and } x_4=1+4(0.5)=3 . \\
& \int_1^3 \sqrt{1+x^2} d x \approx \frac{\Delta x}{2}[f(1)+2 f(1.5)++2 f(2)++2 f(2.5)+f(3)] \\
& =\frac{1}{4}[\sqrt{2}+2 \sqrt{3.25}+2 \sqrt{5}+2 \sqrt{7.25}+\sqrt{10}] \\
& \approx 4.509
\end{aligned}
$
Example2
The function $f$ is continuous on the closed interval $[-1,8]$ and has values that are given in the table above. What is the trapezoidal approximation of $\int_{-1}^8 f(x) d x$ ?
▶️Answer/Explanation
Solution
Make a sketch and use the formula for the area of trapezoid.
$
\begin{aligned}
\text { Area } & =\frac{1}{2}(5+7)(2)+\frac{1}{2}(7+11)(3) \\
& +\frac{1}{2}(11+8)(2)+\frac{1}{2}(8+7)(2) \\
& =12+27+19+15=73
\end{aligned}
$
Example3
- If four equal subdivisions on $[0,2]$ are used, what is the trapezoidal approximation of $\int_0^2 e^x d x$ ?
(A) $\frac{1}{4}\left[1+2 \sqrt{e}+2 e+2 e \sqrt{e}+e^2\right]$
(B) $\frac{1}{2}\left[1+2 \sqrt{e}+2 e+2 e \sqrt{e}+e^2\right]$
(C) $\frac{1}{4}\left[1+\sqrt{e}+e+e \sqrt{e}+e^2\right]$
(D) $\frac{1}{2}\left[1+\sqrt{e}+e+e \sqrt{e}+e^2\right]$
▶️Answer/Explanation
Ans:A
Example4
- If three equal subdivisions on $\left[\frac{\pi}{2}, \pi\right]$ are used, what is the trapezoidal approximation of $\int_{\pi / 2}^\pi \sin x d x ?$
(A) $\frac{\pi}{12}\left(\sin \frac{2 \pi}{3}+\sin \frac{5 \pi}{6}+\sin \pi\right)$
(B) $\frac{\pi}{12}\left(\sin \frac{\pi}{2}+\sin \frac{2 \pi}{3}+\sin \frac{5 \pi}{6}+\sin \pi\right)$
(C) $\frac{\pi}{12}\left(\sin \frac{\pi}{2}+2 \sin \frac{2 \pi}{3}+2 \sin \frac{5 \pi}{6}+\sin \pi\right)$
(D) $\frac{\pi}{6}\left(\sin \frac{\pi}{2}+\sin \frac{2 \pi}{3}+\sin \frac{5 \pi}{6}+\sin \pi\right)$
▶️Answer/Explanation
Ans:C
Example5
- If three equal subdivisions on $[0,6]$ are used, what is the trapezoidal approximation of $\int_0^6 \ln (x+1) d x$ ?
(A) $\frac{1}{3}(\ln 1+\ln 9+\ln 25+\ln 7)$
(B) $\frac{1}{2}(\ln 1+\ln 9+\ln 25+\ln 7)$
(C) $\ln 1+\ln 3+\ln 5+\ln 7$
(D) $\ln 1+\ln 9+\ln 25+\ln 7$
▶️Answer/Explanation
Ans:D
4.6 Fundamental Theorem of Calculus Part 1
If $f$ is continuous on $[a, b]$ then $F(x)=\int_a^x f(t) d t$ is continuous on $[a, b]$ and differentiable on $(a, b)$, so
$
F^{\prime}(x)=\frac{d}{d x} \int_a^x f(t) d t=f(x)
$
If $F(x)=\int_a^{u(x)} f(t) d t$, then by the chain rule
$
F^{\prime}(x)=\frac{d}{d x} \int_a^{u(x)} f(t) d t=f(u) \cdot \frac{d u}{d x}
$
Example1
- If $F(x)=\int_1^x \frac{1}{1+t^3} d t$, then $F^{\prime}(x)=$
▶️Answer/Explanation
Solution
$
\begin{aligned}
& F^{\prime}(x)=\frac{d}{d x} \int_1^x \frac{1}{1+t^3} d t \\
& =\frac{1}{1+x^3}
\end{aligned}
$
Fundamental Theorem of Calculus
Example2
- $
\frac{d}{d x} \int_1^{x^2} \sqrt{3+t^2} d t=
$
(A) $\sqrt{3+x^2}$
(B) $\sqrt{3+x^4}$
(C) $2 x \sqrt{3+x^4}$
(D) $2 \sqrt{3+x^2}$
▶️Answer/Explanation
Ans:B
Example3
- For $-\frac{\pi}{2}<x<\frac{\pi}{2}$, if $F(x)=\int_0^{\sin x} \frac{d t}{\sqrt{1-t^2}}$, then $F^{\prime}(x)=$
(A) $\frac{\sin x}{\sqrt{1-x^2}}$
(B) $\frac{\cos x}{\sqrt{1-x^2}}$
(C) 1
(D) $\csc x$
▶️Answer/Explanation
Ans:C
Example4
- If $F(x)=\int_0^{\sqrt{x}} \cos \left(t^2\right) d t$, then $F^{\prime}(4)=$
(A) $\cos 2$
(B) $\frac{\cos 4}{4}$
(C) $\frac{\cos 4}{\sqrt{2}}$
(D) $\sqrt{2} \cos 4$
▶️Answer/Explanation
Ans:C
4.7 Fundamental Theorem of Calculus Part 2
If $f$ is continuous on $[a, b]$ and $F$ is an antiderivative of $f$ on $[a, b]$, then
$
\int_a^b f(x) d x=F(b)-F(a) .
$
Example1
- $\int_{\pi / 2}^x \cos t d t=$
▶️Answer/Explanation
Solution
$
\text { } \int_{\pi / 2}^x \cos t d t=[\sin t]_{\pi / 2}^x=\left[\sin x-\sin \frac{\pi}{2}\right]=\sin x-1
$
Example 2
- The graph of the function $f$ shown below consists of four line segments. If $g$ is the function defined by $g(x)=\int_{-4}^x f(t) d t$, find the value of $g(6), g^{\prime}(6)$, and $g^{\prime \prime}(6)$.
▶️Answer/Explanation
Solution
$\begin{array}{rlrl}g(6)= & \int_{-4}^6 f(t) d t & & \text { Substitute } 6 \text { for } x . \\ = & \frac{1}{2}(2)(4)-\frac{1}{2}(2)(2)+\frac{1}{2}(2)(4) & & \int_{-4}^6 f(t) d t=\text { sum of the area above } \\ & +(3)(4)+\frac{1}{2}(2+4) \cdot 1 & & \text { the } x \text {-axis minus sum of the area belo } \\ = & 21 & & \text { theaxis, between } x=-4 \text { and } x=6 . \\ g^{\prime}(x) & =f(x) & & \text { Fundamental Theorem of Calculus } \\ g^{\prime}(6) & =f(6)=2 & & \text { Slope of } f \text { at } x=6 \text { is }-2 . \\ g^{\prime \prime}(6) & =f^{\prime}(6)=-2 & \end{array}$
Example 3
- $f$ is the antiderivative of $\frac{\sqrt{x}}{1+x^3}$ such that $f(1)=2$, then $f(3)=$
(A) 1.845
(B) 2.397
(C) 2.906
(D) 3.234
▶️Answer/Explanation
Ans:B
Example 4
- If $f^{\prime}(x)=\cos \left(x^2-1\right)$ and $f(-1)=1.5$, then $f(5)=$
(A) 1.554
(B) 2.841
(C) 3.024
(D) 3.456
▶️Answer/Explanation
Ans:C
Example 5
- If $f(x)=\sqrt{x^4-3 x+4}$ and $g$ is the antiderivative of $f$, such that $g(3)=7$, then $g(0)=$
(A) -2.966
(B) -1.472
(C) -0.745
(D) 1.086
▶️Answer/Explanation
Ans:A
Example 6
- If $f$ is a continuous function and $F^{\prime}(x)=f(x)$ for all real numbers $x$, then $\int_2^{10} f\left(\frac{1}{2} x\right) d x=$
(A) $\frac{1}{2}[F(5)-F(1)]$
(B) $\frac{1}{2}[F(10)-F(2)]$
(C) $2[F(5)-F(1)]$
(D) $2[F(10)-F(2)]$
▶️Answer/Explanation
Ans:B
4.8 Integration by Substitution
The Substitution Rule
If $u=g(x)$ is a differentiable function whose range is an interval $I$ and $f$ is continuous on $I$, then
$
\int f(g(x)) g^{\prime}(x) d x=\int f(u) d u
$
Evaluating the integral $\int f(g(x)) g^{\prime}(x) d x$, when $f$ and $g^{\prime}$ are continuous functions
1. Substitute $u=g(x)$ and $d u=g^{\prime}(x) d x$ to obtain the integral $\int f(u) d u$.
2. Integrate with respect to $u$.
3. Replace $u$ by $g(x)$ in the result.
The Substitution Rule for Definite Integrals
If $g^{\prime}$ is continuous on $[a, b]$ and $f$ is continuous on the range of $u=g(x)$, then
$
\int_a^b f(g(x)) \cdot g^{\prime}(x) d x=\int_{g(a)}^{g(b)} f(u) d u
$
Example1
- Find $\int \cos (5 \theta-3) d \theta$.
▶️Answer/Explanation
Solution
$
\begin{array}{ll}
\int \cos (5 \theta-3) d \theta=\int \cos u \cdot \frac{1}{5} d u & \text { Let } u=5 \theta-3, d u=5 d \theta, \frac{1}{5} d u=d \theta . \\
=\frac{1}{5} \int \cos u d u & \text { The integral is now in standard form. } \\
=\frac{1}{5} \sin u+C & \text { Integrate with respect to } u . \\
=\frac{1}{5} \sin (5 \theta-3)+C & \text { Replace } u \text { by } 5 \theta-3 .
\end{array}
$
Example 2
- Evaluate $\int \frac{x}{\sqrt{1-x^2}} d x$
▶️Answer/Explanation
Solution
$\int \frac{x}{\sqrt{1-x^2}} d x=\int \frac{-1 / 2 d u}{\sqrt{u}}$ Let $u=1-x^2, d u=-2 x d x,-\frac{1}{2} d u=x d x$.
$=-\frac{1}{2} \int u^{-1 / 2} d u$ In the form $\int u^n d u$
$=-\frac{1}{2}(2 \sqrt{u})+C$ Integrate with respect to $u$.
$=-\sqrt{u}+C$
$=-\sqrt{1-x^2}+C$ Replace $u$ with $1-x^2$.
Example 3
- Evaluate $\int_0^{\pi / 2} \frac{3 \cos x}{\sqrt{1+3 \sin x}} d x$
▶️Answer/Explanation
Solution
We have two choices.
Method 1: Transform the integral and evaluate the integral with the transformed limits.
$\int_0^{\pi / 2} \frac{3 \cos x}{\sqrt{1+3 \sin x}} d x$ Let $u=1+3 \sin x, d u=3 \cos x d x$.
When $x=0, u=1+3 \sin 0=1$.
When $x=\pi / 2$,
$\begin{aligned} & =\int_1^4 \frac{d u}{\sqrt{u}} \\ & =\left[2 u^{1 / 2}\right]_1^4 \\ & =2\left[4^{1 / 2}-1\right] \\ & =2[2-1]=2\end{aligned}$ Evaluate the new integral.
Method 2: Transform the integral to an indefinite integral, integrate, change back to $x$, and use the original $x$-limits.
$\int \frac{3 \cos x}{\sqrt{1+3 \sin x}} d x=\int \frac{d u}{\sqrt{u}}$ Let $u=1+3 \sin x, d u=3 \cos x d x$.
$=2 u^{\frac{1}{2}}+C$ Integrate with respect to $u$.
$=2(1+3 \sin x)^{\frac{1}{2}}+C$ Replace $u$ with $1+3 \sin x$.
$\int_0^{\pi / 2} \frac{3 \cos x}{\sqrt{1+3 \sin x}} d x$
$=2\left[(1+3 \sin x)^{\frac{1}{2}}\right]_0^{\pi / 2}$ Use the integral found, with limits of integration for $x$.
$\begin{aligned} & =2\left[\left(1+3 \sin \frac{\pi}{2}\right)^{\frac{1}{2}}-(1+3 \sin 0)^{\frac{1}{2}}\right] \\ & =2\left[4^{\frac{1}{2}}-1^{\frac{1}{2}}\right]=2\end{aligned}$
Example 4
- $
\int \sqrt{x} \sin \left(x^{3 / 2}\right) d x=
$
(A) $\frac{2}{3} \cos \left(x^{3 / 2}\right)+C$
(B) $\frac{2}{3} \sqrt{x} \cos \left(x^{3 / 2}\right)+C$
(C) $-\frac{2}{3} x^{3 / 2} \cos \left(x^{3 / 2}\right)+C$
(D) $-\frac{2}{3} \cos \left(x^{3 / 2}\right)+C$
▶️Answer/Explanation
Ans:D
Example 5
- If the substitution $u=2-x$ is made, $\int_1^3 x \sqrt{2-x} d x=$
(A) $\int_{-1}^1 u \sqrt{u} d u$
(B) $-\int_1^3 u \sqrt{u} d u$
(C) $\int_1^3(2-u) \sqrt{u} d u$
(D) $\int_{-1}^1(u-2) \sqrt{u} d u$
▶️Answer/Explanation
Ans:C
Example 6
- If $\int_{-1}^3 f(x+k) d x=8$, where $k$ is a constant, then $\int_{k-1}^{k+3} f(x) d x=$
(A) $8-k$
(B) $8+k$
(C) 8
(D) $k-8$
▶️Answer/Explanation
Ans:C
4.9 Integration of Exponential and Logarithmic Function
Definition of the Natural Logarithmic Function
The natural logarithmic function is defined by
$
\ln x=\int_1^x \frac{1}{t} d t, \quad x>0
$
If $u$ is a differentiable function such that $u \neq 0$,
$
\int \frac{1}{u} d u=\ln |\boldsymbol{u}|+C
$
Whether $u$ is positive or negative, the integral of $(1 / u) d u$ is $\ln |u|+C$.
Integration of Exponential Function
$
\int e^u d u=e^u+C
$
Example 1
- Evaluate $\int_1^e \frac{x^2+3}{x} d x$
▶️Answer/Explanation
Solution
$\begin{aligned} & \int_1^e \frac{x^2+3}{x} d x=\int_1^e\left(\frac{x^2}{x}+\frac{3}{x}\right) d x=\int_1^e x d x+\int_1^e \frac{3}{x} d x=\left[\frac{x^2}{2}\right]_1^e+[3 \ln x]_1^e \\ & =\left[\frac{e^2}{2}-\frac{1}{2}\right]+3[\ln e-\ln 1]=\frac{e^2}{2}-\frac{1}{2}+3=\frac{e^2}{2}+\frac{5}{2}\end{aligned}$
Example 2
- Evaluate $\int_0^{\pi / 4}\left(e^{\tan x}+2\right) \sec ^2 x d x$.
▶️Answer/Explanation
Solution
$
\int_0^{\pi / 4}\left(e^{\tan x}+2\right) \sec ^2 x d x
$
$=\int_0^1\left(e^u+2\right) d u$ Let $u=\tan x, d u=\sec ^2 x d x$.
When $x=0, u=\tan 0=0$.
When $x=\pi / 4, u=\tan \pi / 4=1$.
$\begin{aligned} & =\left[e^u+2 u\right]_0^1 \\ & =\left[\left(e^1+2\right)-\left(e^0+0\right)\right]=e+1\end{aligned}$
Example 3
- Evaluate $\int_e^{e^2} \frac{(\ln x)^2}{x} d x$.
▶️Answer/Explanation
Solution
$
\int_e^{e^2} \frac{(\ln x)^2}{x} d x
$
$=\int_1^2 u^2 d u$ Let $u=\ln x, d u=d x / x$.
When $x=e, u=\ln e=1$.
When $x=e^2, u=\ln e^2=2$.
$=\left[\frac{1}{3} u^3\right]_1^2=\frac{1}{3}\left[2^3-1^3\right]=\frac{7}{3}$
Example 4
- $
\int_1^3 \frac{x+3}{x^2+6 x} d x=
$
(A) $\ln \frac{3}{2}$
(B) $\frac{\ln 27-\ln 7}{2}$
(C) $\ln 3$
(D) $\frac{\ln 20-\ln 5}{2}$
▶️Answer/Explanation
Ans:B
Example 5
- $
\int_0^1 \frac{x}{e^{x^2}} d x=
$
(A) $e-1$
(B) $\left(1-\frac{1}{e}\right)$
(C) $\frac{1}{2}\left(1-\frac{1}{e}\right)$
(D) $\frac{1}{2}\left(1-\frac{1}{e^2}\right)$
▶️Answer/Explanation
AnsC
Example 6
- $
\int_0^{\pi / 2} \cos x e^{\sin x} d x=
$
(A) $-e$
(B) $1-e$
(C) $\frac{e}{2}$
(D) $e-1$
▶️Answer/Explanation
Ans:D