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AP Calculus AB : Numerical approximations to definite integrals

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Example Question #11 : Numerical Approximations To Definite Integrals

Differentiate $y=3\csc x+5\sin x$

Possible Answers:

$3csc^{2}x-5cosx$

$-3\csc x\cot x+5\cos x$

$3sec^{2}x-5cosx$

$3\csc x\cot x-5\cos x$

$\cot x(3\csc x-5)$

Correct answer:

$-3\csc x\cot x+5\cos x$

Explanation:

The answer is $-3\csc x\cot x+5\cos x$

We simply differentiate by parts, remembering our trig rules.

$y=3\csc x+5\sin x$

$y^{'}= -3\csc x\cot x+5\cos x$

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Example Question #12 : Numerical Approximations To Definite Integrals

If $f(x)=1, f'(x)=2, g(x)=3, \ and\ g'(x)=4$ then find h'(x) .

h(x)=fg

Possible Answers:

Correct answer:

Explanation:

The answer is 10.

h(x)=fg

h'(x)=f'(x)g(x)+f(x)g'(x)

h'(x)=(2)(3)+(1)(4) =10

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Example Question #13 : Numerical Approximations To Definite Integrals

What is the first derivative of the function $\large h(x)=x^{\frac{1}{x}}$ ?

Possible Answers:

$h'(x)=\frac{x^{\frac{1}{x}}}{x^{2}}(1-\ln x)$

$h'(x)=\frac{1-\ln x}{x^{2}}$

$h'(x)=x^{\frac{1}{x}}\ln (x^{2}-1)$

$h'(x)=\frac{1}{x}\cdot x^{\frac{1}{x}-1}$

$h'(x)=\ln (\frac{1}{x})\cdot x^{\frac{1}{x}-1}$

Correct answer:

$h'(x)=\frac{x^{\frac{1}{x}}}{x^{2}}(1-\ln x)$

Explanation:

First, let y=h(x) .

$y=x^{\frac{1}{x}}$

We will take the natural logarithm of both sides in order to simplify the exponential expression on the right.

$\ln y=\ln x^{\frac{1}{x}}$

Next, apply the property of logarithms which states that, in general, $\log x^a=a\log x$ , where is a constant.

$\ln y = \frac{1}{x}\ln x$

We can differentiate both sides with respect to .

$\frac{d}{dx}\[ln y]=\frac{d}{dx}[\frac{1}{x}\ln x]$

We will need to apply the Chain Rule on the left side and the Product Rule on the right side.

$\frac{d}{dy}[\ln y]\cdot \frac{dy}{dx}=\frac{1}{x}\cdot \frac{d}{dx}[\ln x]+\ln x\cdot \frac{d}{dx}[\frac{1}{x}]$

$\frac{1}{y}\cdot \frac{dy}{dx}=\frac{1}{x}\cdot \frac{1}{x} + \ln x\cdot \frac{-1}{x^{2}}$

Because we are looking for the derivative, we must solve for $\frac{\mathrm{d}y }{\mathrm{d} x}$ .

$\frac{dy}{dx}=y\cdot \frac{1}{x^{2}}(1-\ln x)$

However, we want our answer to be in terms of only. We now substitute $x^{\frac{1}{x}}$ in place of .

$\frac{dy}{dx}=\frac{x^{\frac{1}{x}}}{x^{2}}(1-\ln x)$

Since we let y=h(x) , we can replace $\frac{\mathrm{d} y}{\mathrm{d} x}$ with h'(x) .

The answer is $h'(x)=\frac{x^{\frac{1}{x}}}{x^{2}}(1-\ln x)$ .

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Example Question #11 : Numerical Approximations To Definite Integrals

Consider the curve given by the parametric equations below:

$x(t)=\ln(t)$

y(t)=3^t-2^t

What is the equation of the line normal to the curve when t=1 ?

Possible Answers:

$y=\frac{-1}{\ln \frac{27}{4}}x-1$

$y=\frac{-1}{\ln \frac{27}{4}}x+1$

$y=\frac{-1}{\ln \frac{27}{4}}(x-1)$

$y-1={\ln \frac{27}{4}}(x+1)$

$y=\frac{1}{\ln \frac{4}{27}}x+1$

Correct answer:

$y=\frac{-1}{\ln \frac{27}{4}}x+1$

Explanation:

In order to find the equation of the normal line, we will need the slope of the line and a point through which it passes. If we substitute t=1 into our parametric equations, we can easily obtain the point on the curve.

x(1)=0, y(1)=1

The normal line is perpendicular to the tangent line. Thus, we should first find the slope of the tangent line.

To find the value of the tangent slope when t=1 , we will use the following formula:

$\frac{dy}{dx}=\frac{y'(t)}{x'(t)}$

$y'(t)=3^t \ln 3 -2^t \ln 2$

$y'(1)=3\ln3-2\ln 2=\ln \frac{27}{4}$

$x'(t)=\frac{1}{t}$

x'(1)=1

$\frac{dy}{dx}=\frac{y'(t)}{x'(t)}=\ln(\frac{27}{4})$

Because the normal line is perpendicular to the tangent line, the slope of the normal line is the negative reciprocal of the slope of the tangent line. Thus,

slope of normal = $\frac{-1}{\ln(\frac{27}{4})}$ .

We now have the point and slope of the normal line, so we can use point-slope form.

$y-1=\frac{-1}{\ln \frac{27}{4}}(x-0)$

The answer is $y=\frac{-1}{\ln \frac{27}{4}}x+1$ .

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Example Question #1 : Riemann Sums (Left, Right, And Midpoint Evaluation Points)

$\begin{align*}&\text{The function }f(x)\text{ has the following values on the interval }x=[-2,13]:\\&f(-2)=5\\&f(1)=-19\\&f(4)=5\\&f(7)=-6\\&f(10)=-18\\&f(13)=0\\&\\&\text{Approximate the integral of }f(x)\text{ on this interval using right Riemann sums.}\end{align*}$

Possible Answers:

495

Correct answer:

Explanation:

$\begin{align*}&\text{A Riemann sum integral approximation over an interval [a, b]}\\&\text{with n subintervals follows the form:}\\&\int_a^b f(x)dx \approx \frac{b-a}{n}(f(x_1)+f(x_2)+...+f(x_n))\\&\text{It is essentially a sum of n rectangles, each with a base of}\\&\text{length :}\frac{b-a}{n}\text{ and variable heights :}f(x_i)\text{, which depend on the function value at }x_i\\&\text{In our case, we have function values over an interval:}\\&f(-2)=5\\&f(1)=-19\\&f(4)=5\\&f(7)=-6\\&f(10)=-18\\&f(13)=0\\&\\&\text{Although the total interval has a length of }15\\&\text{Notice the points are equidistant. This distance is our subinterval }3\\&\text{and since we are using the right point of each interval, we disregard the first function value:}\\&\int_{-2}^{13}f(x)\approx3[(-19)+(5)+(-6)+(-18)+(0)]\\&\int_{-2}^{13}f(x)\approx-114\end{align*}$

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Example Question #2 : Riemann Sums (Left, Right, And Midpoint Evaluation Points)

$\begin{align*}&\text{Approximate }\int_{-8}^{1}f(x)\\&\text{Using a right Riemann sum, if }f(x)\text{ has the values:}\\&f(-8)=-13\\&f(-5)=-19\\&f(-2)=6\\&f(1)=-9\end{align*}$

Possible Answers:

Correct answer:

Explanation:

$\begin{align*}&\text{A Riemann sum integral approximation over an interval [a, b]}\\&\text{with n subintervals follows the form:}\\&\int_a^b f(x)dx \approx \frac{b-a}{n}(f(x_1)+f(x_2)+...+f(x_n))\\&\text{It is essentially a sum of n rectangles, each with a base of}\\&\text{length :}\frac{b-a}{n}\text{ and variable heights :}f(x_i)\text{, which depend on the function value at }x_i\\&\text{In our case, we have function values over an interval:}\\&f(-8)=-13\\&f(-5)=-19\\&f(-2)=6\\&f(1)=-9\\&\text{Although the total interval has a length of }9\\&\text{Notice the points are equidistant. This distance is our subinterval: }3\\&\text{Since we are using the right point of each interval, we disregard the first function value:}\\&\int_{-8}^{1}f(x)\approx3[(-19)+(6)+(-9)]\\&\int_{-8}^{1}f(x)\approx-66\end{align*}$

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Example Question #3 : Riemann Sums (Left, Right, And Midpoint Evaluation Points)

$\begin{align*}&\text{Approximate }\int_{-2}^{23}f(x)\\&\text{Using a right Riemann sum, if }f(x)\text{ has the values:}\\&f(-2)=-2\\&f(3)=-15\\&f(8)=14\\&f(13)=15\\&f(18)=-9\\&f(23)=-12\end{align*}$

Possible Answers:

Correct answer:

Explanation:

$\begin{align*}&\text{A Riemann sum integral approximation over an interval [a, b]}\\&\text{with n subintervals follows the form:}\\&\int_a^b f(x)dx \approx \frac{b-a}{n}(f(x_1)+f(x_2)+...+f(x_n))\\&\text{It is essentially a sum of n rectangles, each with a base of}\\&\text{length :}\frac{b-a}{n}\text{ and variable heights :}f(x_i)\text{, which depend on the function value at }x_i\\&\text{In our case, we have function values over an interval:}\\&f(-2)=-2\\&f(3)=-15\\&f(8)=14\\&f(13)=15\\&f(18)=-9\\&f(23)=-12\\&\text{Although the total interval has a length of }25\\&\text{Notice the points are equidistant, with }5\text{ intervals between. Each equidistant interval has length: }5\\&\text{Since we are using the right point of each interval, we disregard the first function value:}\\&\int_{-2}^{23}f(x)\approx5[(-15)+(14)+(15)+(-9)+(-12)]\\&\int_{-2}^{23}f(x)\approx-35\end{align*}$

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Example Question #1 : Riemann Sums (Left, Right, And Midpoint Evaluation Points)

$\begin{align*}&\text{Approximate }\int_{-7}^{13}f(x)\\&\text{Using a right Riemann sum, if }f(x)\text{ has the values:}\\&f(-7)=18\\&f(-2)=-17\\&f(3)=-16\\&f(8)=-15\\&f(13)=-14\end{align*}$

Possible Answers:

Correct answer:

Explanation:

$\begin{align*}&\text{A Riemann sum integral approximation over an interval [a, b]}\\&\text{with n subintervals follows the form:}\\&\int_a^b f(x)dx \approx \frac{b-a}{n}(f(x_1)+f(x_2)+...+f(x_n))\\&\text{It is essentially a sum of n rectangles, each with a base of}\\&\text{length :}\frac{b-a}{n}\text{ and variable heights :}f(x_i)\text{, which depend on the function value at }x_i\\&\text{In our case, we have function values over an interval:}\\&f(-7)=18\\&f(-2)=-17\\&f(3)=-16\\&f(8)=-15\\&f(13)=-14\\&\text{Although the total interval has a length of }20\\&\text{, notice the points are equidistant, with }4\text{ intervals between.}\\&\text{Each equidistant interval has length: }5\\&\text{Since we are using the right point of each interval, we disregard the first function value:}\\&\int_{-7}^{13}f(x)\approx5[(-17)+(-16)+(-15)+(-14)]\\&\int_{-7}^{13}f(x)\approx-310\end{align*}$

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Example Question #5 : Riemann Sums (Left, Right, And Midpoint Evaluation Points)

$\begin{align*}&f(4)=9\\&f(6)=10\\&f(8)=-18\\&f(10)=15\\&f(12)=18\\&\text{Approximate, using a right Riemann sum, }\int_{4}^{12}f(x)\end{align*}$

Possible Answers:

272

Correct answer:

Explanation:

$\begin{align*}&\text{A Riemann sum integral approximation over an interval [a, b]}\\&\text{with n subintervals follows the form:}\\&\int_a^b f(x)dx \approx \frac{b-a}{n}(f(x_1)+f(x_2)+...+f(x_n))\\&\text{It is essentially a sum of n rectangles, each with a base of}\\&\text{length :}\frac{b-a}{n}\text{ and variable heights :}f(x_i)\text{, which depend on the function value at }x_i\\&\text{In our case, we have function values over an interval:}\\&f(4)=9\\&f(6)=10\\&f(8)=-18\\&f(10)=15\\&f(12)=18\\&\text{Although the total interval has a length of }8\\&\text{, notice the points are equidistant, with }4\text{ intervals between.}\\&\text{Each equidistant interval has length: }2\\&\text{Since we are using the right point of each interval, we disregard the first function value:}\\&\int_{4}^{12}f(x)\approx2[(10)+(-18)+(15)+(18)]\\&\int_{4}^{12}f(x)\approx50\end{align*}$

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Example Question #6 : Riemann Sums (Left, Right, And Midpoint Evaluation Points)

$\begin{align*}&\text{The function }f(x)\text{ has the following values on the interval }x=[-2,38]:\\&f(-2)=-13\\&f(6)=-4\\&f(14)=-15\\&f(22)=-19\\&f(30)=18\\&f(38)=-8\\&\text{Approximate the integral of }f(x)\text{ on this interval using right Riemann sums.}\end{align*}$

Possible Answers:

Correct answer:

Explanation:

$\begin{align*}&\text{A Riemann sum integral approximation over an interval [a, b]}\\&\text{with n subintervals follows the form:}\\&\int_a^b f(x)dx \approx \frac{b-a}{n}(f(x_1)+f(x_2)+...+f(x_n))\\&\text{It is essentially a sum of n rectangles, each with a base of}\\&\text{length :}\frac{b-a}{n}\text{ and variable heights :}f(x_i)\text{, which depend on the function value at }x_i\\&\text{In our case, we have function values over an interval:}\\&f(-2)=-13\\&f(6)=-4\\&f(14)=-15\\&f(22)=-19\\&f(30)=18\\&f(38)=-8\\&\text{Although the total interval has a length of }40\\&\text{, notice the points are equidistant, with }5\text{ intervals between.}\\&\text{Each equidistant interval has length: }8\\&\text{Since we are using the right point of each interval, we disregard the first function value:}\\&\int_{-2}^{38}f(x)\approx8[(-4)+(-15)+(-19)+(18)+(-8)]\\&\int_{-2}^{38}f(x)\approx-224\end{align*}$

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AP Calculus AB : Numerical approximations to definite integrals

Study concepts, example questions & explanations for AP Calculus AB

All AP Calculus AB Resources

Example Questions

Example Question #11 : Numerical Approximations To Definite Integrals

Example Question #12 : Numerical Approximations To Definite Integrals

Example Question #13 : Numerical Approximations To Definite Integrals

Example Question #11 : Numerical Approximations To Definite Integrals

Example Question #1 : Riemann Sums (Left, Right, And Midpoint Evaluation Points)

Example Question #2 : Riemann Sums (Left, Right, And Midpoint Evaluation Points)

Example Question #3 : Riemann Sums (Left, Right, And Midpoint Evaluation Points)

Example Question #1 : Riemann Sums (Left, Right, And Midpoint Evaluation Points)

Example Question #5 : Riemann Sums (Left, Right, And Midpoint Evaluation Points)

Example Question #6 : Riemann Sums (Left, Right, And Midpoint Evaluation Points)

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