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Calculus 1 : Functions

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Example Questions

Example Question #51 : Midpoint Riemann Sums

Using the method of midpoint Reimann sums, approximate the integral of the function $f(x)=e^{\frac{1}{x}}$ $\displaystyle f(x)=e^{\frac{1}{x}}$ over the interval x=[1,6] $\displaystyle x=[1,6]$ using four midpoints.

Possible Answers:

7.1815 $\displaystyle 7.1815$

5.1283 $\displaystyle 5.1283$

8.6178 $\displaystyle 8.6178$

5.7452 $\displaystyle 5.7452$

10.9372 $\displaystyle 10.9372$

Correct answer:

7.1815 $\displaystyle 7.1815$

Explanation:

The Reimann sum approximation of an integral of a function with $\displaystyle n$ subintervals over an interval [a,b] $\displaystyle [a,b]$ takes the form:

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

Where $\frac{b-a}{n}$ $\displaystyle \frac{b-a}{n}$ is the length of the subintervals.

For this problem, since there are four midpoints, the subintervals have length $\frac{6-1}{4}=1.25$ $\displaystyle \frac{6-1}{4}=1.25$, and the midpoints are $\displaystyle x=[1.625,2.875,4.125,5.375]$.

The integral is thus:

$\int_1^6 e^{\frac{1}{x}}dx \approx 1.25[e^{\frac{1}{1.625}}+e^{\frac{1}{2.875}}+e^{\frac{1}{4.125}}+e^{\frac{1}{5.375}}]$ $\displaystyle \int_1^6 e^{\frac{1}{x}}dx \approx 1.25[e^{\frac{1}{1.625}}+e^{\frac{1}{2.875}}+e^{\frac{1}{4.125}}+e^{\frac{1}{5.375}}]$

$\int_1^6 e^{\frac{1}{x}}dx \approx 7.1815$ $\displaystyle \int_1^6 e^{\frac{1}{x}}dx \approx 7.1815$

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Example Question #1081 : Calculus

Using the method of midpoint Reimann sums, approximate the area between the curves $f(x)=e^{\frac{1}{x}}$ $\displaystyle f(x)=e^{\frac{1}{x}}$ and $g(x)=e^{x^2}$ $\displaystyle g(x)=e^{x^2}$ over the interval $\displaystyle x=[0.5,1]$ using four midpoints.

Possible Answers:

0.6214 $\displaystyle 0.6214$

2.4571 $\displaystyle 2.4571$

0.8952 $\displaystyle 0.8952$

1.1314 $\displaystyle 1.1314$

1.5927 $\displaystyle 1.5927$

Correct answer:

1.1314 $\displaystyle 1.1314$

Explanation:

The Reimann sum approximation of an integral of a function with $\displaystyle n$ subintervals over an interval [a,b] $\displaystyle [a,b]$ takes the form:

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

Where $\frac{b-a}{n}$ $\displaystyle \frac{b-a}{n}$ is the length of the subintervals.

For this problem, since there are four midpoints, the subintervals have length $\frac{1-0.5}{4}=0.125$ $\displaystyle \frac{1-0.5}{4}=0.125$, and the midpoints are $\displaystyle x=[0.5675,0.6925,0.8175,0.9325]$.

Since this problem is dealing with the area of the region between functions, it's essentially asking for the difference of the larger and smaller areas. Over the given interval, $e^{\frac{1}{x}}>e^{x^2}$ $\displaystyle e^{\frac{1}{x}}>e^{x^2}$.

The integral is thus:

${\int_{0.5}^1 (e^{\frac{1}{x}}-e^{x^2})dx \approx 0.125[(e^{\frac{1}{0.5675}}-e^{0.5675^2})+(e^{\frac{1}{0.6925}}-e^{0.6925^2})+(e^{\frac{1}{0.8175}}-e^{0.8175^2})+(e^{\frac{1}{0.9325}}-e^{0.9325^2})]}$ $\displaystyle {\int_{0.5}^1 (e^{\frac{1}{x}}-e^{x^2})dx \approx 0.125[(e^{\frac{1}{0.5675}}-e^{0.5675^2})+(e^{\frac{1}{0.6925}}-e^{0.6925^2})+(e^{\frac{1}{0.8175}}-e^{0.8175^2})+(e^{\frac{1}{0.9325}}-e^{0.9325^2})]}$

${\int_{0.5}^1 (e^{\frac{1}{x}}-e^{x^2})dx \approx 1.1314$ $\displaystyle {\int_{0.5}^1 (e^{\frac{1}{x}}-e^{x^2})dx \approx 1.1314$

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Example Question #1082 : Calculus

Possible Answers:

49.7709 $\displaystyle 49.7709$

36.9908 $\displaystyle 36.9908$

6.2213 $\displaystyle 6.2213$

12.4427 $\displaystyle 12.4427$

24.8854 $\displaystyle 24.8854$

Correct answer:

12.4427 $\displaystyle 12.4427$

Explanation:

The Reimann sum approximation of an integral of a function with $\displaystyle n$ subintervals over an interval [a,b] $\displaystyle [a,b]$ takes the form:

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

Where $\frac{b-a}{n}$ $\displaystyle \frac{b-a}{n}$ is the length of the subintervals.

For this problem, since there are four midpoints, the subintervals have length $\frac{2-1}{4}=0.25$ $\displaystyle \frac{2-1}{4}=0.25$, and the midpoints are $\displaystyle x=[1.125,1.375,1.625,1.875]$.

Since this problem is dealing with the area of the region between functions, it's essentially asking for the difference of the larger and smaller areas. Over the given interval, $e^{\frac{1}{x}}<e^{x^2}$ $\displaystyle e^{\frac{1}{x}}< e^{x^2}$.

The integral is thus:

${\int_1^2 (e^{x^2}-e^{\frac{1}{x}})dx \approx 0.25[(e^{1.125^2}-e^{\frac{1}{1.125}})+(e^{1.375^2}-e^{\frac{1}{1.375}})+(e^{1.625^2}-e^{\frac{1}{1.625}})+(e^{1.875^2}-e^{\frac{1}{1.875}})]}$ $\displaystyle {\int_1^2 (e^{x^2}-e^{\frac{1}{x}})dx \approx 0.25[(e^{1.125^2}-e^{\frac{1}{1.125}})+(e^{1.375^2}-e^{\frac{1}{1.375}})+(e^{1.625^2}-e^{\frac{1}{1.625}})+(e^{1.875^2}-e^{\frac{1}{1.875}})]}$

${\int_1^2 (e^{x^2}-e^{\frac{1}{x}})dx \approx 12.4427$ $\displaystyle {\int_1^2 (e^{x^2}-e^{\frac{1}{x}})dx \approx 12.4427$

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Example Question #1083 : Calculus

The velocity of an errant particle is given by the function $v(t)=\frac{1}{2}e^{tsin(t)}$ $\displaystyle v(t)=\frac{1}{2}e^{tsin(t)}$. Approximate the average velocity of the particle over the interval of time $t=[0,4\pi]$ $\displaystyle t=[0,4\pi]$ using the method of midpoint Reimann sums and four midpoints.

Possible Answers:

1118.3 $\displaystyle 1118.3$

2136.7 $\displaystyle 2136.7$

659.2 $\displaystyle 659.2$

322.6 $\displaystyle 322.6$

4053.9 $\displaystyle 4053.9$

Correct answer:

322.6 $\displaystyle 322.6$

Explanation:

Average velocity can be found as total distance traveled divided by total time elapsed.

Distance traveled over an interval of time can be found by integrating the velocity function over this interval of time. For this problem, Reimann sums will be used to approximate this integral.

The Reimann sum approximation of an integral of a function with $\displaystyle n$ subintervals over an interval [a,b] $\displaystyle [a,b]$ takes the form:

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

Where $\frac{b-a}{n}$ $\displaystyle \frac{b-a}{n}$ is the length of the subintervals.

For this problem, since there are four midpoints, the subintervals have length $\frac{4\pi-0}{4}=\pi$ $\displaystyle \frac{4\pi-0}{4}=\pi$, and the midpoints are $x=[\frac{\pi}{2},\frac{3\pi}{2},\frac{5\pi}{2},\frac{7\pi}{2}]$ $\displaystyle x=[\frac{\pi}{2},\frac{3\pi}{2},\frac{5\pi}{2},\frac{7\pi}{2}]$.

The integral is thus:

$\int_0^{4\pi} f(x)dx \approx \pi[(\frac{1}{2}e^{\frac{\pi}{2}sin(\frac{\pi}{2})})+(\frac{1}{2}e^{\frac{3\pi}{2}sin(\frac{3\pi}{2})})+(\frac{1}{2}e^{\frac{5\pi}{2}sin(\frac{5\pi}{2})})+(\frac{1}{2}e^{\frac{7\pi}{2}sin(\frac{7\pi}{2})})]$ $\displaystyle \int_0^{4\pi} f(x)dx \approx \pi[(\frac{1}{2}e^{\frac{\pi}{2}sin(\frac{\pi}{2})})+(\frac{1}{2}e^{\frac{3\pi}{2}sin(\frac{3\pi}{2})})+(\frac{1}{2}e^{\frac{5\pi}{2}sin(\frac{5\pi}{2})})+(\frac{1}{2}e^{\frac{7\pi}{2}sin(\frac{7\pi}{2})})]$

$\int_0^{4\pi} f(x)dx \approx 4053.9$ $\displaystyle \int_0^{4\pi} f(x)dx \approx 4053.9$

Now note that this is an approximation of the distance traveled and we want the average velocity. To find this, divide this value by the time elapsed, which is $4\pi$ $\displaystyle 4\pi$:

$v_{avg}=322.6$ $\displaystyle v_{avg}=322.6$

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Example Question #1084 : Calculus

Using the method of midpoint Reimann sums, approximate the integral $\int_3^5 x^{\frac{1}{x}}dx$ $\displaystyle \int_3^5 x^{\frac{1}{x}}dx$ using five midpoints.

Possible Answers:

10.233 $\displaystyle 10.233$

6.281 $\displaystyle 6.281$

1.413 $\displaystyle 1.413$

7.991 $\displaystyle 7.991$

2.827 $\displaystyle 2.827$

Correct answer:

2.827 $\displaystyle 2.827$

Explanation:

The Reimann sum approximation of an integral of a function with $\displaystyle n$ subintervals over an interval [a,b] $\displaystyle [a,b]$ takes the form:

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

Where $\frac{b-a}{n}$ $\displaystyle \frac{b-a}{n}$ is the length of the subintervals.

For this problem, since there are five midpoints, the subintervals have length $\frac{5-3}{5}=0.4$ $\displaystyle \frac{5-3}{5}=0.4$, and the midpoints are $\displaystyle x=[3.2,3.6,4.0,4.4,4.8]$.

This gives the integral approximation:

$\int_3^5 x^{\frac{1}{x}}dx \approx 0.4[3.2^{\frac{1}{3.2}}+3.6^{\frac{1}{3.6}}+4.0^{\frac{1}{4.0}}+4.4^{\frac{1}{4.4}}+4.8^{\frac{1}{4.8}}]$ $\displaystyle \int_3^5 x^{\frac{1}{x}}dx \approx 0.4[3.2^{\frac{1}{3.2}}+3.6^{\frac{1}{3.6}}+4.0^{\frac{1}{4.0}}+4.4^{\frac{1}{4.4}}+4.8^{\frac{1}{4.8}}]$

$\int_3^5 x^{\frac{1}{x}}dx \approx 2.827$ $\displaystyle \int_3^5 x^{\frac{1}{x}}dx \approx 2.827$

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Example Question #51 : Functions

Using the method of midpoint Reimann sums, approximate the integral $\int_0^1 x^{x^x} dx$ $\displaystyle \int_0^1 x^{x^x} dx$ using four midpoints.

Possible Answers:

1.150 $\displaystyle 1.150$

0.575 $\displaystyle 0.575$

3.451 $\displaystyle 3.451$

2.301 $\displaystyle 2.301$

1.726 $\displaystyle 1.726$

Correct answer:

0.575 $\displaystyle 0.575$

Explanation:

The Reimann sum approximation of an integral of a function with $\displaystyle n$ subintervals over an interval [a,b] $\displaystyle [a,b]$ takes the form:

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

Where $\frac{b-a}{n}$ $\displaystyle \frac{b-a}{n}$ is the length of the subintervals.

For this problem, since there are four midpoints, the subintervals have length $\frac{1-0}{4}=0.25$ $\displaystyle \frac{1-0}{4}=0.25$, and the midpoints are $\displaystyle x=[0.125,0.375,0.625,0.875]$.

This gives the integral approximation:

$\int_0^1 x^{x^x} dx \approx 0.25[0.125^{0.125^{0.125}}+0.375^{0.375^{0.375}}+0.625^{0.625^{0.625}}+0.875^{0.875^{0.875}}]$ $\displaystyle \int_0^1 x^{x^x} dx \approx 0.25[0.125^{0.125^{0.125}}+0.375^{0.375^{0.375}}+0.625^{0.625^{0.625}}+0.875^{0.875^{0.875}}]$

$\int_0^1 x^{x^x} dx \approx 0.575$ $\displaystyle \int_0^1 x^{x^x} dx \approx 0.575$

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Example Question #51 : Midpoint Riemann Sums

Approximate the integral $\int_1^4 2^{2^{x}}dx$ $\displaystyle \int_1^4 2^{2^{x}}dx$ using the method of midpoint Reimann sums and three midpoints.

Possible Answers:

2603.01 $\displaystyle 2603.01$

276 $\displaystyle 276$

2851.91 $\displaystyle 2851.91$

65842.5 $\displaystyle 65842.5$

65808 $\displaystyle 65808$

Correct answer:

2603.01 $\displaystyle 2603.01$

Explanation:

The Reimann sum approximation of an integral of a function with $\displaystyle n$ subintervals over an interval [a,b] $\displaystyle [a,b]$ takes the form:

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

Where $\frac{b-a}{n}$ $\displaystyle \frac{b-a}{n}$ is the length of the subintervals.

For this problem, since there are three midpoints, the subintervals have length $\frac{4-1}{3}=1$ $\displaystyle \frac{4-1}{3}=1$, and the midpoints are $\displaystyle x=[1.5,2.5,3.5]$.

Thus this gives the integral approximation:

$\int_1^4 2^{2^{x}}dx \approx [2^{2^{1.5}}+2^{2^{2.5}}+2^{2^{3.5}}]$ $\displaystyle \int_1^4 2^{2^{x}}dx \approx [2^{2^{1.5}}+2^{2^{2.5}}+2^{2^{3.5}}]$

$\int_1^4 2^{2^{x}}dx \approx 2603.01$ $\displaystyle \int_1^4 2^{2^{x}}dx \approx 2603.01$

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Example Question #51 : Differential Functions

Approximate the integral $\int_{-\pi}^{\pi}x^{sin(x)}dx$ $\displaystyle \int_{-\pi}^{\pi}x^{sin(x)}dx$ using the method of midpoint Reimann sums and four midpoints.

Possible Answers:

2.16i $\displaystyle 2.16i$

$\displaystyle 2.56-2.16i$

-2.16i $\displaystyle -2.16i$

2.56 $\displaystyle 2.56$

$\displaystyle 2.56+2.16i$

Correct answer:

$\displaystyle 2.56-2.16i$

Explanation:

The Reimann sum approximation of an integral of a function with $\displaystyle n$ subintervals over an interval [a,b] $\displaystyle [a,b]$ takes the form:

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

Where $\frac{b-a}{n}$ $\displaystyle \frac{b-a}{n}$ is the length of the subintervals.

For this problem, since there are four midpoints, the subintervals have length $\frac{\pi-(-\pi)}{4}=\frac{\pi}{2}$ $\displaystyle \frac{\pi-(-\pi)}{4}=\frac{\pi}{2}$, and the midpoints are $x=[\frac{-3\pi}{4},\frac{-\pi}{4},\frac{\pi}{4},\frac{3\pi}{4}]$ $\displaystyle x=[\frac{-3\pi}{4},\frac{-\pi}{4},\frac{\pi}{4},\frac{3\pi}{4}]$.

This gives the integral approximation:

$\int_{-1}^{1}x^{sin(x)}dx \approx \frac{\pi}{2}[\frac{-3\pi}{4}^{sin(\frac{-3\pi}{4})}+\frac{-\pi}{4}^{sin(\frac{-\pi}{4})}+\frac{\pi}{4}^{sin(\frac{\pi}{4})}+\frac{3\pi}{4}^{sin(\frac{3\pi}{4})}]$ $\displaystyle \int_{-1}^{1}x^{sin(x)}dx \approx \frac{\pi}{2}[\frac{-3\pi}{4}^{sin(\frac{-3\pi}{4})}+\frac{-\pi}{4}^{sin(\frac{-\pi}{4})}+\frac{\pi}{4}^{sin(\frac{\pi}{4})}+\frac{3\pi}{4}^{sin(\frac{3\pi}{4})}]$

$\int_{-1}^{1}x^{sin(x)}dx \approx 2.56-2.16i$ $\displaystyle \int_{-1}^{1}x^{sin(x)}dx \approx 2.56-2.16i$

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Example Question #51 : Midpoint Riemann Sums

Approximate the integral $\int_5^8 x^{\frac{1}{x}}dx$ $\displaystyle \int_5^8 x^{\frac{1}{x}}dx$ using the method of midpoint Reimann sums and two midpoints.

Possible Answers:

4.005 $\displaystyle 4.005$

1.127 $\displaystyle 1.127$

4.926 $\displaystyle 4.926$

1.335 $\displaystyle 1.335$

2.670 $\displaystyle 2.670$

Correct answer:

4.005 $\displaystyle 4.005$

Explanation:

The Reimann sum approximation of an integral of a function with $\displaystyle n$ subintervals over an interval [a,b] $\displaystyle [a,b]$ takes the form:

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

Where $\frac{b-a}{n}$ $\displaystyle \frac{b-a}{n}$ is the length of the subintervals.

For this problem, since there are two midpoints, the subintervals have length $\frac{8-5}{2}=1.5$ $\displaystyle \frac{8-5}{2}=1.5$, and the midpoints are $\displaystyle x=[5.75,7.25]$.

The integral approximation becomes:

$\int_5^8 x^{\frac{1}{x}}dx \approx 1.5[5.75^{\frac{1}{5.75}}+7.25^{\frac{1}{7.25}}]$ $\displaystyle \int_5^8 x^{\frac{1}{x}}dx \approx 1.5[5.75^{\frac{1}{5.75}}+7.25^{\frac{1}{7.25}}]$

$\int_5^8 x^{\frac{1}{x}}dx \approx 4.005$ $\displaystyle \int_5^8 x^{\frac{1}{x}}dx \approx 4.005$

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Example Question #52 : Midpoint Riemann Sums

Approximate the integral $\int_2^6\frac{1}{x}^{\frac{1}{x}}dx$ $\displaystyle \int_2^6\frac{1}{x}^{\frac{1}{x}}dx$ using the method of midpoint Reimann sums and two midpoints.

Possible Answers:

1.299 $\displaystyle 1.299$

2.898 $\displaystyle 2.898$

3.123 $\displaystyle 3.123$

1.418 $\displaystyle 1.418$

2.836 $\displaystyle 2.836$

Correct answer:

2.836 $\displaystyle 2.836$

Explanation:

The Reimann sum approximation of an integral of a function with $\displaystyle n$ subintervals over an interval [a,b] $\displaystyle [a,b]$ takes the form:

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

Where $\frac{b-a}{n}$ $\displaystyle \frac{b-a}{n}$ is the length of the subintervals.

For this problem, since there are two midpoints, the subintervals have length $\frac{6-2}{2}=2$ $\displaystyle \frac{6-2}{2}=2$, and the midpoints are x=[3,5] $\displaystyle x=[3,5]$.

This gives the integral approximation

$\int_2^6\frac{1}{x}^{\frac{1}{x}}dx \approx 2[\frac{1}{3}^{\frac{1}{3}}+\frac{1}{5}^{\frac{1}{5}}]$ $\displaystyle \int_2^6\frac{1}{x}^{\frac{1}{x}}dx \approx 2[\frac{1}{3}^{\frac{1}{3}}+\frac{1}{5}^{\frac{1}{5}}]$

$\int_2^6\frac{1}{x}^{\frac{1}{x}}dx \approx 2.836$ $\displaystyle \int_2^6\frac{1}{x}^{\frac{1}{x}}dx \approx 2.836$

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Calculus 1 : Functions

Study concepts, example questions & explanations for Calculus 1

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Example Questions

Example Question #51 : Midpoint Riemann Sums

Example Question #1081 : Calculus

Example Question #1082 : Calculus

Example Question #1083 : Calculus

Example Question #1084 : Calculus

Example Question #51 : Functions

Example Question #51 : Midpoint Riemann Sums

Example Question #51 : Differential Functions

Example Question #51 : Midpoint Riemann Sums

Example Question #52 : Midpoint Riemann Sums

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