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Calculus 2 : Calculus II

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9 Diagnostic Tests 308 Practice Tests Question of the Day Flashcards Learn by Concept

Example Questions

Example Question #1 : Riemann Sum: Midpoint Evaluation

Approximate

$\int_{1}^{5} \frac{ \ln x \; \mathrm {d}x }{x^2}$

using the midpoint rule with n=4 . Round your answer to three decimal places.

Possible Answers:

0.382

0.828

0.414

0.503

0.446

Correct answer:

0.503

Explanation:

The interval $\left [ 1,5 \right ]$ is 4 units in width; the interval is divided evenly into four subintervals $\Delta x = 4\div 4 = 1$ units in width, with their midpoints shown:

$\left [ 1,2 \right ] : x _{1} = 1.5$

$\left [2,3 \right ]: x _{2} = 2.5$

$\left [3,4 \right ] : x _{3} = 3.5$

$\left [ 4,5 \right ] : x _{4} = 4.5$

The midpoint rule requires us to calculate:

$M = \Delta x \left [ f (x_{1} )+ f (x_{2}) + f (x_{3}) + f (x_{4}) \right ]$

where $\Delta x = 1$ and $f(x) = \frac{ \ln x }{x^2}$

Evaluate $f(x) = \frac{ \ln x }{x^2}$ for each of $x = x_ { 1 }, x_ { 2 }, x_ { 3 }, x_ { 4 }$ :

$f(1.5) = \frac{ \ln 1.5 }{1.5^2} \approx \frac{0.4055}{2.25} \approx 0.1802$

$f(2.5) = \frac{ \ln 2.5 }{2.5^2} \approx \frac{0.9163}{6.25} \approx 0.1466$

$f(3.5) = \frac{ \ln 3.5 }{3.5^2} \approx \frac{1.2528}{12.25} \approx 0.1023$

$f(4.5) = \frac{ \ln 4.5 }{4.5^2} \approx \frac{1.5041}{20.25} \approx 0.0743$

$M =1 \left (0.1802 + 0.1466 +0.1023 + 0.0743 \right )$

= 0.503

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Example Question #1 : Riemann Sum: Midpoint Evaluation

Approximate

$\int_{1}^{2} x^{2} \ln x \; \mathrm {d}x$

using the midpoint rule with n = 5 . Round your answer to three decimal places.

Possible Answers:

Cannot be determined

1.064

0.806

1.360

1.083

Correct answer:

1.064

Explanation:

The interval $\left [ 1,2 \right ]$ is 1 unit in width; the interval is divided evenly into five subintervals $\Delta x = 1 \div 5 = 0.2$ units in width, with their midpoints shown:

$\left [ 1, 1.2 \right ]; x_{1} = 1.1$

$\left [ 1.2, 1.4 \right ]; x_{2} = 1.3$

$\left [1.4, 1.6 \right ]; x_{3} =1.5$

$\left [ 1.6, 1.8 \right ]; x_{4} = 1.7$

$\left [1.8, 2 \right ];x_{5} = 1.9$

The midpoint rule requires us to calculate:

$M = \Delta x \left ( f (x_{1} )+ f (x_{2}) + f (x_{3}) + f (x_{4}) + f (x_{5}) \right )$

where $\Delta x = 0.2$ and $f(x) =x^{2} \ln x$

Evaluate $f(x) =x^{2} \ln x$ for each of $x = x_ { 1 }, x_ { 2 }, x_ { 3 }, x_ { 4 }, x_ { 5 }$ :

$f(1.1) = 1.1 ^{2} \cdot \ln 1.1 \approx 1.1 ^{2} \cdot 0.0953 \approx 0.1153$

$f(1.3) = 1.3^{2} \cdot \ln 1.3 \approx 1.3 ^{2} \cdot 0.2624 \approx 0.4434$

$f(1.5) = 1.5 ^{2} \cdot \ln 1.5 \approx 1.5 ^{2} \cdot 0.4055 \approx 0.9123$

$f(1.7) = 1.7 ^{2} \cdot \ln 1.7\approx 1.7 ^{2} \cdot 0.5306 \approx 1.5335$

$f(1.9) = 1.9 ^{2} \cdot \ln 1.9 \approx 1.9 ^{2} \cdot 0.6419 \approx 2.3171$

$M \approx 0.2 \cdot \left ( 0.1153 +0.4434 + 0.9123 + 1.5335 + 2.3171 \right ) \approx 1.064$

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Example Question #1 : Trapezoidal Sums

Approximate

$\int_{1}^{2} e ^{x^{2}} \mathrm{d}x$

using the trapezoidal rule with n=5 . Round your answer to three decimal places.

Possible Answers:

20.878

10.502

15.690

14.644

31.379

Correct answer:

15.690

Explanation:

The interval $\left [ 1,2 \right ]$ is 1 unit in width; the interval is divided evenly into five subintervals $\Delta x = 1 \div 5 = 0.2$ units in width. They are

$\left [ 1,1.2 \right ], \left [ 1.2, 1.4 \right ], \left [ 1.4, 1.6 \right ], \left [ 1.6, 1.8 \right ], \left [ 1.8, 2 \right ]$ .

The trapezoidal rule approximates the area of the given integral $\int_{1}^{2} e ^{x^{2}} \mathrm{d}x$ by evaluating

$T = \frac{\Delta x}{2} \left [ f(x_{0}) + 2 f(x_{1}) + 2 f(x_{2}) + 2 f(x_{3}) + 2 f(x_{4}) + f(x_{5}) \right ]$ ,

where

$\Delta x = 0.2$

$f (x) = e ^{x^{2}}$

and

$x_{0} = 1, x_{1} = 1.2,...x_{5} = 2$ .

$T =\frac{0.2}{2} \left [ f(1) + 2 f(1.2}) + 2 f(1.4) + 2 f(1.6) + 2 f(1.8) + f(2) \right ]$

$f(1) = e ^{1^{2}} \approx 2.7183$

$f(1.2) = e ^{1.2^{2}} \approx 4.2207$

$f(1.4) = e ^{1.4^{2}} \approx 7.0993$

$f(1.6) = e ^{1.6^{2}} \approx 12.9358$

$f(1.8) = e ^{1.8^{2}} \approx 25.5337$

$f(2) = e ^{2^{2}} \approx 54.5982$

$T \approx \frac{0.2}{2} \left [2.7183 + 2 \cdot 4.2207 + 2 \cdot 7.0993 + 2 \cdot 12.9358 + 2 \cdot 25.5337 + 54.5982 \right ]$

$\approx 0.1 \left (2.7183 + 8.4414 + 14.1986 + 25.8716 + 51.0674 + 54.5982 \right )$

$\approx 0.1 \cdot 156.8955 \approx 15.690$

Report an Error

Example Question #42 : Integrals

Approximate

$\int_{0}^{ \frac{\pi}{4}} \tan ^{2}x \; \mathrm{d}x$

using the trapezoidal rule with n = 4 . Round your answer to three decimal places.

Possible Answers:

0.325

0.258

0.454

0.129

0.227

Correct answer:

0.227

Explanation:

The interval $\left [ 0, \frac{\pi }{4}\right ]$ is $\frac{\pi }{4}$ units in width; the interval is divided evenly into four subintervals $\Delta x = \frac{\pi }{4} \div 4 = \frac{\pi }{16}$ units in width. They are

$\left [ 0, \frac{\pi }{16} \right ], \left [ \frac{\pi }{16}, \frac{\pi }{8} \right ], \left [ \frac{\pi }{8}, \frac{3 \pi }{16} \right ], \left [ \frac{3 \pi }{16}, \frac{\pi }{4}\right ]$ .

The trapezoidal rule approximates the area of the given integral $\int_{0}^{ \frac{\pi}{4}} \tan ^{2}x \; \mathrm{d}x$ by evaluating

$T = \frac{\Delta x}{2} \left [ f(x_{0}) + 2 f(x_{1}) + 2 f(x_{2}) + 2 f(x_{3}) + 2 f(x_{4}) + f(x_{5}) \right ]$ ,

where

$\Delta x = \frac{\pi }{16}$ ,

$f (x) = \tan ^{2}x$ ,

and

$x_{0} = 0, x_{1} = \frac{\pi }{16},...x_{4} = \frac{\pi }{4}$ .

$T =\frac{\frac{\pi }{16}}{2} \left [ f(0) + 2 f \left ( \frac{ \pi }{16}\right ) + 2 f \left ( \frac{ \pi }{8}\right ) + 2 f \left ( \frac{ 3 \pi }{16}\right ) + f \left ( \frac{ \pi }{4} \right ) \right ]$

$f(0) = \tan ^{2} 0 = 0$

$f\left ( \frac{ \pi }{16}\right ) = \tan ^{2} \frac{ \pi }{16} \approx 0.0396$

$f\left ( \frac{ \pi }{8}\right ) = \tan ^{2} \frac{ \pi }{8} \approx 0.1716$

$f\left ( \frac{ 3 \pi }{16}\right ) = \tan ^{2} \frac{ 3 \pi }{16} \approx 0.4465$

$f\left ( \frac{ \pi }{4}\right ) = \tan ^{2} \frac{ \pi }{4} = 1$

$\frac{\frac{\pi }{16}}{2} = \frac{\pi }{32} = 0.0982$

$T \approx 0.0982 \left [0+ 2 \cdot 0.0396 + 2 \cdot 0.1716 + 2 \cdot 0.4465+ 1 \right ]$

$\approx 0.0982 \left [0+ 0.0792 + 0.3432 + 0.8930 + 1 \right ]$

$\approx 0.0982 \cdot 2.3154 \approx 0.227$

Report an Error

Example Question #43 : Integrals

Approximate

$\int_{1}^{5} \frac{ \ln x \; \mathrm {d}x }{x^2}$

using the trapezoidal rule with n = 4 . Round your estimate to three decimal places.

Possible Answers:

0.503

0.382

0.446

0.828

0.414

Correct answer:

0.414

Explanation:

The interval $\left [ 1,5 \right ]$ is 4 units in width; the interval is divided evenly into four subintervals $\Delta x = 4\div 4 = 1$ units in width - they are [1,2], [2,3],[3,4], [4,5] .

The trapezoidal rule approximates the area of the given integral $\int_{1}^{5} \frac{ \ln x \; \mathrm {d}x }{x^2}$ by evaluating

$T = \frac{\Delta x}{2} \left [ f(x_{0}) + 2 f(x_{1}) + 2 f(x_{2}) + 2 f(x_{3}) + f(x_{4}) \right ]$ ,

where $\Delta x = 1$ , $f(x) = \frac{ \ln x }{x^2}$ , and

$x_{0} = 1, x_{1} = 2, ... x_{4} = 5$ .

$T =\frac{1}{2} \left [ f(1) + 2 f(2}) + 2 f(3) + 2 f(4) + f(5) \right ]$

$f(1) = \frac{ \ln 1 }{1^2} = 0$

$f(2) = \frac{ \ln 2 }{2^2} \approx \frac{0.6931 }{4} \approx 0.1733$

$f(3) = \frac{ \ln 3 }{3^2} \approx \frac{1.0986 }{9} \approx 0.1221$

$f(4) = \frac{ \ln 4 }{4^2} \approx \frac{1.3863 }{16} \approx 0.0866$

$f(5) = \frac{ \ln 5 }{5^2} \approx \frac{1.6094 }{25} \approx 0.0644$

$T \approx 0.5 \left [0+ 2 \cdot 0.1733 + 2 \cdot 0.1221 + 2 \cdot 0.0866 + 0.0644 \right ]$

$\approx 0.5 \left [ 0.3466 + 0.2442 + 0.1732 + 0.0644 \right ]$

$\approx 0.5 \cdot 0.8284 \approx 0.414$

Report an Error

Example Question #2 : Riemann Sum: Midpoint Evaluation

Approximate

$\int_{0}^{\frac{\pi}{2}} x^{2} \cos x \; \mathrm {d}x$

using the midpoint rule with n = 5 . Round your answer to three decimal places.

Possible Answers:

0.470

0.478

0.467

0.447

None of the other choices are correct.

Correct answer:

0.478

Explanation:

The interval $\left [ 0, \frac{\pi}{2}\right ]$ is $\frac{\pi}{2}$ units in width; the interval is divided evenly into five subintervals $\Delta x = \frac{\pi}{2} \div 5 = \frac{\pi}{10}$ units in width, with their midpoints shown:

$\left [ 0, \frac{\pi}{10}\right ]; x_{1} = \frac{\pi}{20}$

$\left [ \frac{\pi}{10},\frac{\pi}{5} \right ]; x_{2} = \frac{3 \pi}{20}$

$\left [ \frac{\pi}{5},\frac{3\pi}{10} \right ]; x_{3} =\frac{5 \pi}{20} = \frac{ \pi}{4}$

$\left [ \frac{3\pi}{10}, \frac{2 \pi}{5} \right ]; x_{4} =\frac{7 \pi}{20}$

$\left [ \frac{2 \pi}{5} , \frac{\pi}{2} \right ];x_{5} = \frac{9 \pi}{20}$

The midpoint rule requires us to calculate:

$M = \Delta x \left ( f (x_{1} )+ f (x_{2}) + f (x_{3}) + f (x_{4}) + f (x_{5}) \right )$

where $\Delta x = \frac{\pi}{10}$ and $f(x) = x^{2} \cos x$

Evaluate $f(x) = x^{2} \cos x$ for each of $x = x_ { 1 }, x_ { 2 }, x_ { 3 }, x_ { 4 }, x_ { 5 }$ :

$f\left $ \frac{\pi}{20} \right$ = \left $ \frac{\pi}{20} \right$^{2} \cos \left $ \frac{\pi}{20} \right$$

$\approx 0.1571^{2} \cdot \cos 0.1571 \approx 0.1571^{2} \cdot 0.9877 \approx 0.0244$

$f\left $ \frac{3\pi}{20} \right$ = \left $ \frac{3\pi}{20} \right$^{2} \cos \left $ \frac{3\pi}{20} \right$$

$\approx 0.4712 ^{2} \cdot \cos 0.4712 \approx 0.1571^{2} \cdot 0.8910 \approx 0.1979$

$f\left $ \frac{\pi}{4} \right$ = \left $ \frac{\pi}{4} \right$^{2} \cos \left $ \frac{\pi}{4} \right$$

$\approx 0.7854^{2} \cdot \cos 0.7854 \approx 0.7854^{2} \cdot 0.7071 \approx 0.4362$

$f\left $ \frac{7\pi}{20} \right$ = \left $ \frac{7\pi}{20} \right$^{2} \cos \left $ \frac{7\pi}{20} \right$$

$\approx 1.0996^{2} \cdot \cos 1.0996 \approx 1.0996^{2} \cdot 0.4540 \approx 0.5489$

$f\left $ \frac{9\pi}{20} \right$ = \left $ \frac{9\pi}{20} \right$^{2} \cos \left $ \frac{9\pi}{20} \right$$

$\approx 1.4137 ^{2} \cdot \cos1.4137 \approx 1.4137^{2} \cdot 0.1564 \approx 0.3126$

Since $\Delta x = \frac{\pi}{10} \approx 0.3142$ ,

we can approximate as

$M \approx 0.3142 \cdot \left ( 0.0244 + 0.1979 + 0.4362 + 0.5489 + 0.3126) \right ) \approx 0.478$ .

Report an Error

Example Question #44 : Integrals

Using Simpson's parabolic rule with n = 4 , give an approximation of

$\int_{0}^{4} \ln (\cosh x) \; \mathrm{d}x$ .

Round your approximation to the nearest thousandth.

Possible Answers:

5.643

7.375

5.597

5.722

4.068

Correct answer:

5.643

Explanation:

The interval [0,4] is divided into four subintervals of width $\Delta x = 1$ by the numbers $\left\{0,1,2,3,4 \right \}$ .

Simpson's parabolic rule tells us that we can approximate

$\int_{0}^{4} \ln (\cosh x) \; \mathrm{d}x$

with n = 4 using the formula

$\frac{\Delta x}{3} \left ( f(x_{0}) +4 f(x_{1}) +2f(x_{2})+4f(x_{3}) +f(x_{4}) \right )$ ,

where

$\Delta x = 1$ ,

$f(x) = \ln (\cosh x)$ ,

and

$x_{0}= 0, x_{1} = 1...x_{4} = 4$ .

Therefore,

$\int_{0}^{4} \ln (\cosh x) \; \mathrm{d}x$

$\approx \frac{1}{3} \left ( f(0) +4 f(1) +2f(2)+4f(3) +f(4) \right )$ .

Evaluate at each of these values of :

$f(0) = \ln \left ( \cosh 0 \right ) = \ln 1 = 0$

$f(1) = \ln \left ( \cosh 1 \right ) \approx \ln 1.5431 \approx 0.4338$

$f(2) = \ln \left ( \cosh 2 \right ) \approx \ln 3.7622 \approx 1.3250$

$f(3) = \ln \left ( \cosh 3 \right ) \approx \ln 10.0677 \approx 2.3093$

$f(4) = \ln \left ( \cosh 4 \right ) \approx \ln 27.3082 \approx 3.3072$

Substitute:

$\frac{1}{3} \left (0 +4 \cdot 0.4338 +2 \cdot 1.3250 +4 \cdot 2.3093 +3.3072 \right )$

$\approx \frac{1}{3} \left (0 +1.7352 +2.6500 +9.2372 +3.3072 \right )$

$\approx \frac{1}{3} \left (16.9296 \right ) \approx 5.643$

Report an Error

Example Question #43 : Integrals

Using Simpson's parabolic rule with n = 4 , give an approximation of

$\int_{0}^{1} 2^{\sin \pi x} \; \mathrm{d}x$ .

Round your approximation to the nearest thousandth.

Possible Answers:

1.566

1.422

1.588

1.755

1.600

Correct answer:

1.588

Explanation:

The interval [0,1] is divided into four subintervals of width $\Delta x = \frac{1}{4}$ by the numbers $\left\{0, \frac{1}{4}, \frac{1}{2}, \frac{3}{4},1 \right \}$ .

Simpson's parabolic rule tells us that we can approximate

$\int_{0}^{1} 2^{\sin \pi x} \; \mathrm{d}x$

with n = 4 using the formula

$\frac{\Delta x}{3} \left ( f(x_{0}) +4 f(x_{1}) +2f(x_{2})+4f(x_{3}) +f(x_{4}) \right )$ ,

where

$\Delta x = \frac{1}{4}$ ,

$f(x) = 2^{\sin \pi x}$ ,

and

$x_{0}= 0, x_{1} = \frac{1}{4}...x_{4} = 1$ .

Therefore,

$\int_{0}^{1} 2^{\sin \pi x} \; \mathrm{d}x$

$\approx \frac{\frac{1}{4}}{3} \left ( f(0) +4 f \left ( \frac{1}{4} \right ) +2f\left ( \frac{1}{2} \right )+4f\left ( \frac{3}{4} \right ) +f(1) \right )$

$\approx \frac{1}{12} \left ( f(0) +4 f \left ( \frac{1}{4} \right ) +2f\left ( \frac{1}{2} \right )+4f\left ( \frac{3}{4} \right ) +f(1) \right )$

Evaluate at each of these values of :

$f(0) = 2^{\sin\left ( \pi \cdot 0 \right )} = 2^{ 0} = 1$

$f\left ( \frac{1}{4} \right ) = 2^{\sin\left ( \pi \cdot \frac{1}{4} \right )} = 2^{\sin \frac{\pi }{4} }= 2^{ \frac{\sqrt{2 }}{2} } \approx 1.6325$

$f\left ( \frac{1}{2} \right ) = 2^{\sin\left ( \pi \cdot \frac{1}{2} \right )} = 2^{\sin \frac{\pi }{2} }= 2^{1}= 2$

$f\left ( \frac{3}{4} \right ) = 2^{\sin\left ( \pi \cdot \frac{3}{4} \right )} = 2^{\sin \frac{3\pi }{4} }= 2^{ \frac{\sqrt{2 }}{2} } \approx 1.6325$

$f(1) = 2^{\sin \left (\pi \cdot 1 \right )} = 2^{\sin \pi } = 2^{ 0} = 1$

Substitute:

$\frac{1}{12} \left ( f(0) +4 f \left ( \frac{1}{4} \right ) +2f\left ( \frac{1}{2} \right )+4f\left ( \frac{3}{4} \right ) +f(1) \right )$

$\approx \frac{1}{12} \left ( 1 +4 \cdot 1.6325 +2 \cdot 2+4\cdot 1.6325 +1 \right )$

$\approx \frac{1}{12} \left ( 1 +6.5300 +4+6.5300 +1 \right )$

$\approx \frac{1}{12} \cdot 19.0600 \approx 1.588$

Report an Error

Example Question #45 : Introduction To Integrals

Using Simpson's parabolic rule with n = 4 , give, to the nearest thousandth, an approximation of

$\int_{1}^{3} \ln (x^{2}+x + 1) \; \mathrm{d} x$ .

Possible Answers:

4.173

3.816

3.807

3.821

3.440

Correct answer:

3.816

Explanation:

The interval [1,3] is divided into four subintervals of width $\Delta x = 0.5$ by the numbers $\left\{1, 1.5, 2, 2.5,3 \right \}$ .

Simpson's parabolic rule tells us that we can approximate

$\int_{1}^{3} \ln (x^{2}+x + 1) \; \mathrm{d} x$

with n = 4 using the formula

$\frac{\Delta x}{3} \left ( f(x_{0}) +4 f(x_{1}) +2f(x_{2})+4f(x_{3}) +f(x_{4}) \right )$ ,

where

$\Delta x = 0.5$ ,

$f(x) = \ln (x^{2}+x + 1)$ ,

and

$x_{0}= 1, x_{1} =1.5,...x_{4} = 3$ .

Therefore,

$\int_{1}^{3} \ln (x^{2}+x + 1) \; \mathrm{d} x$

$= \frac{0.5}{3} \left ( f(1) +4 f(1.5) +2f(2)+4f(2.5) +f(3) \right )$

$= \frac{1}{6} \left ( f(1) +4 f(1.5) +2f(2)+4f(2.5) +f(3) \right )$

Evaluate for each of these values, leaving the results in logarithmic form for the time being:

$f(1) = \ln (1^{2}+1 + 1) = \ln 3$

$f(1.5) = \ln (1.5^{2}+1.5 + 1) = \ln 4.75$

$f(2) = \ln (2^{2}+2 + 1) = \ln 7$

$f(2.5) = \ln (2.5^{2}+2.5 + 1) = \ln 9.75$

$f(3) = \ln (3^{2}+3 + 1) = \ln 13$

$\frac{1}{6} \left ( f(1) +4 f(1.5) +2f(2)+4f(2.5) +f(3) \right )$

$= \frac{1}{6} \left ( \ln 3 +4\ln 4.75 +2 \ln 7+4 \ln 9.75 +\ln 13 \right )$

$= \frac{1}{6} \ln \left (3 \cdot 4.75^{4} \cdot 7^{2}\cdot 9.75^{4} \cdot 13 \right )$

$= \ln \sqrt[6]{3 \cdot 4.75^{4} \cdot 7^{2}\cdot 9.75^{4} \cdot 13 }$

$\approx \ln 45.4300 \approx 3.816$

Report an Error

Example Question #1 : Riemann Sums

Find the Left Riemann sum of the function

f(x)=x+7

on the interval [0,8] divided into four sub-intervals.

Possible Answers:

100

Correct answer:

Explanation:

The interval [0, 8] divided into four sub-intervals gives rectangles with vertices of the bases at

x=0,2,4,6,8

For the Left Riemann sum, we need to find the rectangle heights which values come from the left-most function value of each sub-interval, or f(0), f(2), f(4), and f(6).

f(0) = 0+7 = 7

f(2) = 2+7 = 9

f(4) = 4+7 = 11

f(6) = 6+7 = 13

Because each sub-interval has a width of 2, the Left Riemann sum is

A = bh = 2(7+9+11+13) = 80

Report an Error

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Calculus 2 : Calculus II

Study concepts, example questions & explanations for Calculus 2

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Example Questions

Example Question #1 : Riemann Sum: Midpoint Evaluation

Example Question #1 : Riemann Sum: Midpoint Evaluation

Example Question #1 : Trapezoidal Sums

Example Question #42 : Integrals

Example Question #43 : Integrals

Example Question #2 : Riemann Sum: Midpoint Evaluation

Example Question #44 : Integrals

Example Question #43 : Integrals

Example Question #45 : Introduction To Integrals

Example Question #1 : Riemann Sums

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