Functions, Graphs, and Limits

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AP Calculus AB › Functions, Graphs, and Limits

Questions 1 - 10

Evaluate the following indefinite integral.

$\int (1-x)dx$

$x-\frac{1}{2}x^2+C$

$x+\frac{1}{2}x^2+C$

Explanation

Use the inverse Power Rule to evaluate the integral. We know that $\int x^n dx= \frac{x^{n+1}}{n+1}+C$ for $n \ne -1$ . We see that this rule tells us to increase the power of by 1 and multiply by $\frac{1}{n+1}$ Next always add your constant of integration that would be lost in the differentiation. Take the derivative of your answer to check your work.

Evaluate the following indefinite integral.

$\int (4x^7+x^4+x)dx$

$\frac{1}{2}x^8+\frac{1}{5}x^5+\frac{1}{2}x^2+C$

${2}x^8+{5}x^5+{2}x^2+C$

$\frac{1}{7}x^7+\frac{1}{4}x^4+x+C$

$\frac{1}{2}x^8+\frac{1}{5}x^4+\frac{1}{2}x^2$

Explanation

Use the inverse Power Rule to evaluate the integral. We know that $\int x^n dx= \frac{x^{n+1}}{n+1}+C$ for $n \ne -1$ . We see that this rule tells us to increase the power of by 1 and multiply by $\frac{1}{n+1}$ . Next always add your constant of integration that would be lost in the differentiation. Take the derivative of your answer to check your work.

Evaluate the following indefinite integral.

$\int 4xdx$

Explanation

First, we know that we can pull the constant "4" out of the integral, and we then evaluate the integral according to this equation:

$\int x^n dx=\frac{x^{n+1}}{n+1}+C, n\ne1$ . From this, we acquire the answer above. As a note, we cannot forget the constant of integration which would be lost during the differentiation.

Asymptoteplot

$\begin{align*}&\text{Decide which of the following functions, most likely represents the sample data above.}\end{align*}$

$-\frac{(x - 11)}{(4x + 1)}$

$-\frac{(4x + 1)}{(x - 11)}$

$-\frac{(x + 11)}{(4x - 1)}$

$-\frac{(4x - 1)}{(x + 11)}$

Explanation

$\begin{align*}&\text{Observation of the data points shows that there’s a sharp increase}\&\text{and decrease on either side of a particular x-value. This type}\&\text{of behavior is observed when there’s a vertical asymptote. An}\&\text{asymptote is a value that a function may approach, but will}\&\text{never actually attain. In the case of vertical asymptotes, this}\&\text{behavior occurs if the function approaches infinity for a given}\&\text{x-value, often when a zero value appears in a denominator. Noting}\&\text{this rule, the above function has a zero in the denominator}\&\text{at a definite point centered approximately around:}\&x=-0.3\&\text{We find a zero denominator for the function: }\&-\frac{(x - 11)}{(4x + 1)}\end{align*}$

Evaluate the following indefinite integral.

$\int \frac{1}{2}xdx$

$\frac{1}{4}x^2+C$

$\frac{1}{2}x^2+C$

$\frac{1}{2}x+C$

Explanation

Use the inverse Power Rule to evaluate the integral. Firstly, constants can be taken out of the integral, so we pull the 1/2 out front and then complete the integration according to the rule. We know that $\int x^n dx= \frac{x^{n+1}}{n+1}+C$ for $n \ne -1$ . We see that this rule tells us to increase the power of by 1 and multiply by $\frac{1}{n+1}$ . Next always add your constant of integration that would be lost in the differentiation. Take the derivative of your answer to check your work.

Find the indefinite ingtegral for $\int\frac{x+1}{\sqrt{x}} dx$ .

$\frac{2}{3}x^{\frac{3}{2}}+2x^{\frac{1}{2}}+C$

$x^{}\frac{3}{2}+ x^{\frac{1}{2}}+C$

$(x-2)^\frac{3}{2}$

Explanation

First, bring up the radical into the numerator and distribute to the (x+1) term.

$(x+1)x^{\frac{-1}{2}}$

$\int x^{\frac{1}{2}} + x^{\frac{-1}{2}}dx$

Then integrate.

$\frac{2}{3}x^{\frac{3}{2}}+ 2x^{\frac{1}{2}}$

Since it's indefinite, don't forget to add the C:

$\frac{2}{3}x^{\frac{3}{2}}+ 2x^{\frac{1}{2}}+C$

Asymptoteplot

$\begin{align*}&\text{Decide which of the following functions, most likely represents the sample data above.}\end{align*}$

$-\frac{(x - 11)}{(4x + 1)}$

$-\frac{(4x + 1)}{(x - 11)}$

$-\frac{(x + 11)}{(4x - 1)}$

$-\frac{(4x - 1)}{(x + 11)}$

Explanation

Asymptoteplot

$\begin{align*}&\text{Of the following functions, which most likely represents the sample data above?}\end{align*}$

$-\frac{(10x + 11)}{(11x + 20)}$

$-\frac{(11x + 20)}{(10x + 11)}$

$-\frac{(10x - 11)}{(11x - 20)}$

$-\frac{(11x - 20)}{(10x - 11)}$

Explanation

$\begin{align*}&\text{Observation of the data points shows that there’s a sharp increase}\&\text{and decrease on either side of a particular x-value. This type}\&\text{of behavior is observed when there’s a vertical asymptote. An}\&\text{asymptote is a value that a function may approach, but will}\&\text{never actually attain. In the case of vertical asymptotes, this}\&\text{behavior occurs if the function approaches infinity for a given}\&\text{x-value, often when a zero value appears in a denominator. Noting}\&\text{this rule, the above function has a zero in the denominator}\&\text{at a definite point centered approximately around:}\&x=-1.8\&\text{We find a zero denominator for the function: }\&-\frac{(10x + 11)}{(11x + 20)}\end{align*}$

Asymptoteplot

$\begin{align*}&\text{Which of the four following functions is probably depicted in the sample data above?}\end{align*}$

$-\frac{(19x - 15)}{(5x - 15)}$

$-\frac{(5x - 15)}{(19x - 15)}$

$\frac{(19x - 15)}{(5x - 15)}$

$\frac{(5x - 15)}{(19x - 15)}$

Explanation

$\begin{align*}&\text{Notice that as x increases, the points of data graphed appears}\&\text{to level out, flattening towards a certain value. This value}\&\text{is what is known as a horizontal asymptote. An asymptote is}\&\text{a value that a function approaches, but never actually reaches.}\&\text{Think of a horizontal asymptote as the limit of a function as}\&\text{x approaches infinity. In such a case, as x approaches infinity,}\&\text{any constants added or subtracted in the numerator and denominator}\&\text{become irrelevant. What matters is the power of x in the denominator}\&\text{and the numerator; if those are the same, then the coefficients}\&\text{define the asymptote. We see that the function flattens towards:}\&f(\infty)=-3.8\&\text{This matches the ratio of coefficients for the function: }\&-\frac{(19x - 15)}{(5x - 15)}\end{align*}$

Evaluate the following indefinite integral.

$\int \frac{3}{x}dx$

$\frac{1}{3}ln|x|+C$

$\frac{3}{2}x^3+C$

Explanation

Use the inverse Power Rule to evaluate the integral. We know that $\int x^n dx= \frac{x^{n+1}}{n+1}+C$ for $n \ne -1$ . But, in this case, IS equal to so a special condition of the rule applies. We must instead use $\int x^{-1}dx=ln|x|+C$ . Pull the constant "3" out front and evaluate accordingly. Next always add your constant of integration that would be lost in the differentiation. Take the derivative of your answer to check your work.

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