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Calculus 1 : Functions

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Example Questions

Example Question #21 : Differential Functions

Solve the integral

$\int_{0}^{9}{\frac{1}{1+e^{x}}}dx$

using Simpson's rule with 2n=10 subintervals.

Possible Answers:

0.3482

0.8829

1.2369

0.6599

0.6924

Correct answer:

0.6924

Explanation:

Simpson's rule is solved using the formula

$\int_{a}^{b}f(x))dx\approx S_{2m}=\frac{1}{3}(\frac{b-a}{2n})\left [ f(0)+4f(1)+2f(2)+...+2f(m-2)+4f(m-1)+f(m) \right ]$

where is the number of subintervals and f(m) is the function evaluated at the midpoint.

For this problem, $(\frac{b-a}{2n})=(\frac{9-0}{10})=0.9$ .

The value of each approximation term is below.

Screen shot 2015 06 11 at 9.36.20 pm

The sum of all the approximation terms is 2.3081 therefore

$(\frac{b-a}{2n})\left [ f(0)+2f(1)+...+f(m-1)+f(m) \right ]=(0.9)*(2.3081)=0.6924$

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Example Question #22 : Differential Functions

Approximate the area under the curve from x=0 to x=4 for the function

f(x)=x^3-2x+2

using midpoint Riemann sum when n=4 .

Possible Answers:

108

Correct answer:

Explanation:

The find the area under the curve for the function you use the formula $\int_{a}^{b}f(x)dx$ .

Riemann's midpoint formula states that

$\int_{a}^{b}f(x)dx= \frac{b-a}{n}(f(x_{1})+f(x_{2})+f(x_{3})+...+f(x_{n}))$ .

This method breaks the area under the curve into n rectangles. The first term of the equation that is multiplied out find the base length of each rectangle and then the f(x) terms are the heights at the middle of each rectangle. By multiplying the two terms, you find the area and when you add them together you find the approximation of the area under the curve.

We will plug in our values into this formula. a=0, b=4 and n=4. To find the x values to plug into the function we look at the base length of each rectangle.

$\frac{b-a}{n}=\frac{4-0}{4}=1$

so there is a rectangle wall every x=1, thus the midpoints will be x1=0.5, x2=1.5, x3=2.5, x4=3.5.

$\\ \int_{a}^{b}f(x)dx= \frac{4-0}{4}(f(0.5)+f(1.5)+f(2.5)+f(3.5))\\ = 1(1.125+2.375+12.625+37.875)\\ =54.$

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Example Question #23 : Differential Functions

Approximate the $\int_{1}^{3}\frac{4x+3}{2}dx$ using midpoint Riemann sum when n=4 .

Possible Answers:

6.5

Correct answer:

Explanation:

The find the area under the curve for the function you use the formula $\int_{a}^{b}f(x)dx$ .

Riemann's midpoint formula states that

$\int_{a}^{b}f(x)dx= \frac{b-a}{n}(f(x_{1})+f(x_{2})+f(x_{3})+...+f(x_{n}))$ .

We will plug in our values into this formula. a=1, b=3 and n=4. To find the x values to plug into the function we look at the base length of each rectangle.

$\frac{b-a}{n}=\frac{3-1}{4}=0.5$

So there is a rectangle wall every x=0.5 starting at x=1, thus the midpoints will be x1=1.25, x2=1.75, x3=2.25, x4=2.75.

$\\ \int_{a}^{b}f(x)dx= \frac{3-1}{4}(f(1.25)+f(1.75)+f(2.25)+f(2.75))\\ = 0.5(4+5+6+7)\\=11$

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Example Question #24 : Differential Functions

Approximate the $\int_{1}^{9}\frac{10}{x^2+2}dx$ using midpoint Riemann sum when n=4 .

Possible Answers:

5.28

-3.21

2.64

9.66

Correct answer:

5.28

Explanation:

The find the area under the curve for the function you use the formula $\int_{a}^{b}f(x)dx$ .

Riemann's midpoint formula states that

$\int_{a}^{b}f(x)dx= \frac{b-a}{n}(f(x_{1})+f(x_{2})+f(x_{3})+...+f(x_{n}))$ .

We will plug in our values into this formula. a=1, b=9 and n=4. To find the x values to plug into the function we look at the base length of each rectangle.

$\frac{b-a}{n}=\frac{9-1}{4}=2$

so there is a rectangle wall every x=2 starting at x=1, thus the midpoints will be x1=2, x2=4, x3=6, x4=8.

$\\ \int_{a}^{b}f(x)dx= \frac{9-1}{4}(f(2)+f(4)+f(6)+f(8))\\ = 2\left(\frac{5}{3}+\frac{5}{9}+\frac{5}{19}+\frac{5}{33}\right)\\ \approx 5.28$

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Example Question #21 : Midpoint Riemann Sums

Approximate the integral $\int_{0}^{5}e^{\frac{x^2}{25}}dx$ using midpoint Reimann sums and four midpoints.

Possible Answers:

7.244

8.527

6.823

6.379

5.795

Correct answer:

7.244

Explanation:

The form of a Riemann sum follows:

$\int_{a}^{b}f(x)dx\approx \frac{b-a}{n}(f(x_1)+f(x_2)+...+f(x_n))$

The interval [0,5] can be divided into four intervals of length $\frac{5-0}{4}=1.25$

[0,1.25][1.25,2.5][2.5,3.75][3.75,5]

Each of these intervals in turn have respective midpoints :

[0.625,1.875,3.125,4.375]

Therefore, the approximation of the integral is:

$\frac{5-0}{4}(e^{\frac{0.625^2}{25}}+e^{\frac{1.875^2}{25}}+e^{\frac{3.125^2}{25}}+e^{\frac{4.375^2}{25}})$

7.244

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Example Question #21 : Functions

Find the midpoint Reimann sum for the integral of the function f(x)=2x over the interval x=[4,14] using five midpoints.

Possible Answers:

160

200

180

Correct answer:

180

Explanation:

The form of a Riemann sum follows:

$\int_{a}^{b}f(x)dx\approx \frac{b-a}{n}(f(x_1)+f(x_2)+...+f(x_n))$

Five intervals can be made of length $\frac{14-4}{5}=2$ with midpoints [5,7,9,11,13]

Thus, the Reimann sum is:

$\frac{14-4}{5}(2(5)+2(7)+2(9)+2(11)+2(13))$

180

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Example Question #21 : Functions

Using the method of midpoint Reimann sums, approximate the integral $\int_{1}^{4}x^xdx$ using three midpoints.

Possible Answers:

287

275.793

91.931

Correct answer:

91.931

Explanation:

The form of a Riemann sum follows:

$\int_{a}^{b}f(x)dx\approx \frac{b-a}{n}(f(x_1)+f(x_2)+...+f(x_n))$

The interval [1,4] can be divided into intervals of length $\frac{4-1}{3}=1$ with midpoints [1.5,2.5,3.5] .

Therefore, the integral approximation is:

$\frac{4-1}{3}(1.5^{1.5}+2.5^{2.5}+3.5^{3.5})$

91.931

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Example Question #28 : Differential Functions

Using midpoint Riemann sums, approximate the integral of x^3 over the interval [0,4] with four points.

Possible Answers:

42.875

248

Correct answer:

Explanation:

To solve for the midpoint Riemann sum, the function must be evaluated at x=0.5,1.5,2.5, and 3.5.

y(0.5)=0.5^3=0.125

y(1.5)=1.5^3=3.375

y(2.5)=2.5^3=15.625

y(3.5)=3.5^3=42.875

These must each be multiplied by the segment inteval

$= \frac{total interval}{number of segments }= \frac{4}{4} =1$ .

These are then summed to yield the midpoint Riemann sum approximation:

$0.125\cdot 1+3.375\cdot 1+15.625\cdot 1+42.875\cdot 1=62$

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Example Question #29 : Differential Functions

Use the Midpoint Reimann Sum approximation method to approximate the area between [1,3] using four subintervals under the curve y=x^2+10 .

Possible Answers:

28.5

26.825

28.625

Correct answer:

28.625

Explanation:

This approximation method has a few steps. First find out how wide your rectangles are. Our interval is [1,3] and we need 4 subintervals.

$\frac{(3-1)}{4}=.5$ tells us that each subinterval is 0.5.

Now we need to figure out high each rectangle is. We're going to be finding those y-values at the midpoints of each triangle; namely, at x=1.25, 1.75, 2.25, and 2.75.

We will then plug those x-values into the function given and then multiply by the width of our rectangles (0.5).

Therefore, our equation looks like:

0.5(11.5625+13.0625+15.0625+17.5625) .

This gives us an area of 28.625 .

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Example Question #30 : Differential Functions

Use midpoint Riemann sums to approximate the area between the x-axis and the curve $f(x)=\frac{1}{ln(x^2)}$ for x=2 to , using three midpoints.

Possible Answers:

2.554

1.537

2.253

1.277

1.126

Correct answer:

1.277

Explanation:

The form of a Riemann sum follows:

$\int_{a}^{b}f(x)dx\approx \frac{b-a}{n}(f(x_1)+f(x_2)+...+f(x_n))$

For $f(x)=\frac{1}{ln(x^2)}$ for x=2 to , our three midpoints are [2.5,3.5,4.5] .

So the approximation of the area under the curve will be:

$\frac{5-2}{3}\left(\frac{1}{ln(2.5^2)}+\frac{1}{ln(3.5^2)}+\frac{1}{ln(4.5^2)}\right)$

1.277

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Calculus 1 : Functions

Study concepts, example questions & explanations for Calculus 1

All Calculus 1 Resources

Example Questions

Example Question #21 : Differential Functions

Example Question #22 : Differential Functions

Example Question #23 : Differential Functions

Example Question #24 : Differential Functions

Example Question #21 : Midpoint Riemann Sums

Example Question #21 : Functions

Example Question #21 : Functions

Example Question #28 : Differential Functions

Example Question #29 : Differential Functions

Example Question #30 : Differential Functions

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