Trapezoidal Sums - AP Calculus BC

Card 0 of 6

Question

Approximate

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

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

Answer

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

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

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

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

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

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

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

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

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

So

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

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

$\approx 0.1 \cdot 156.8955 \approx 15.690$

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Question

Approximate

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

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

Answer

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

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

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

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

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

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

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

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

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

So

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

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

$\approx 0.0982 \cdot 2.3154 \approx 0.227$

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Question

Approximate

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

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

Answer

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 .

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

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

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

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

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

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

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

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

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

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

$\approx 0.5 \cdot 0.8284 \approx 0.414$

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Question

Approximate

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

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

Answer

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

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

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

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

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

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

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

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

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

So

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

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

$\approx 0.1 \cdot 156.8955 \approx 15.690$

Compare your answer with the correct one above

Question

Approximate

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

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

Answer

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

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

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

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

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

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

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

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

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

So

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

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

$\approx 0.0982 \cdot 2.3154 \approx 0.227$

Compare your answer with the correct one above

Question

Approximate

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

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

Answer

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 .

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

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

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

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

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

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

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

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

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

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

$\approx 0.5 \cdot 0.8284 \approx 0.414$

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