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High School Math : Calculus II — Integrals

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Example Questions

Example Question #1 : Understanding Maclaurin Series

Give the $x^{4}$ $\displaystyle x^{4}$ term of the Maclaurin series expansion of the function $f(x) = x \tan^{-1} x$ $\displaystyle f(x) = x \tan^{-1} x$.

Possible Answers:

$- \frac{1}{12} x^{4 }$ $\displaystyle - \frac{1}{12} x^{4 }$

$- \frac{1}{4} x^{4 }$ $\displaystyle - \frac{1}{4} x^{4 }$

$\displaystyle 0$

$- \frac{3}{4} x^{4 }$ $\displaystyle - \frac{3}{4} x^{4 }$

$- \frac{1}{3} x^{4 }$ $\displaystyle - \frac{1}{3} x^{4 }$

Correct answer:

$- \frac{1}{3} x^{4 }$ $\displaystyle - \frac{1}{3} x^{4 }$

Explanation:

This can most easily be answered by recalling that the Maclaurin series for $\tan^{-1} x$ $\displaystyle \tan^{-1} x$ is

$\tan^{-1} x = x - \frac{1}{3} x^{3 } + \frac{1}{5} x^{5 }...$ $\displaystyle \tan^{-1} x = x - \frac{1}{3} x^{3 } + \frac{1}{5} x^{5 }...$

Multiply by $\displaystyle x$ to get:

$x\tan^{-1} x = x\cdot x - x\cdot \frac{1}{3} x^{3 } + x\cdot \frac{1}{5} x^{5 }...$ $\displaystyle x\tan^{-1} x = x\cdot x - x\cdot \frac{1}{3} x^{3 } + x\cdot \frac{1}{5} x^{5 }...$

$x \tan^{-1} x=x^{2 } - \frac{1}{3} x^{4 } + \frac{1}{5} x^{6 }...$ $\displaystyle x \tan^{-1} x=x^{2 } - \frac{1}{3} x^{4 } + \frac{1}{5} x^{6 }...$

The $x^{4}$ $\displaystyle x^{4}$ term is therefore $- \frac{1}{3} x^{4 }$ $\displaystyle - \frac{1}{3} x^{4 }$.

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Example Question #2 : Understanding Maclaurin Series

Give the $x^ {3}$ $\displaystyle x^ {3}$ term of the Maclaurin series of the function $f(x) = \sinh 3x$ $\displaystyle f(x) = \sinh 3x$ .

Possible Answers:

$\frac{9}{2} x^{3}$ $\displaystyle \frac{9}{2} x^{3}$

$-9x^{3}$ $\displaystyle -9x^{3}$

$9x^{3}$ $\displaystyle 9x^{3}$

$-\frac{9}{2} x^{3}$ $\displaystyle -\frac{9}{2} x^{3}$

$\displaystyle 0$

Correct answer:

$\frac{9}{2} x^{3}$ $\displaystyle \frac{9}{2} x^{3}$

Explanation:

The $x^ {3}$ $\displaystyle x^ {3}$ term of a Maclaurin series expansion has coefficient

$\frac{f ' '' (0)}{3!} = \frac{f ' '' (0)}{6}$ $\displaystyle \frac{f ' '' (0)}{3!} = \frac{f ' '' (0)}{6}$.

We can find f'''(x) $\displaystyle f'''(x)$ by differentiating three times in succession:

$f(x) = \sinh 3x$ $\displaystyle f(x) = \sinh 3x$

$f'(x) = 3 \cosh 3x$ $\displaystyle f'(x) = 3 \cosh 3x$

$f''(x) = 3\cdot 3\sinh 3x = 9\sinh 3x$ $\displaystyle f''(x) = 3\cdot 3\sinh 3x = 9\sinh 3x$

$f'''(x) = 9 \cdot 3\cosh 3x = 27 \cosh 3x$ $\displaystyle f'''(x) = 9 \cdot 3\cosh 3x = 27 \cosh 3x$

$f'''(0) = 27 \cosh \left (3 \cdot 0 \right ) = 27 \cosh 0 = 27\cdot 1 = 27$ $\displaystyle f'''(0) = 27 \cosh \left (3 \cdot 0 \right ) = 27 \cosh 0 = 27\cdot 1 = 27$

The term we want is therefore

$\frac{f ' '' (0)}{6} \cdot x^{3} = \frac{27}{6} \cdot x^{3} = \frac{9}{2} x^{3}$ $\displaystyle \frac{f ' '' (0)}{6} \cdot x^{3} = \frac{27}{6} \cdot x^{3} = \frac{9}{2} x^{3}$

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Example Question #2251 : High School Math

Give the $x^{2}$ $\displaystyle x^{2}$ term of the Maclaurin series expansion of the function $f (x) = 7 ^{x}$ $\displaystyle f (x) = 7 ^{x}$.

Possible Answers:

$\frac{\left ( \ln 7 \right )^{2}}{2} x^{2}$ $\displaystyle \frac{\left ( \ln 7 \right )^{2}}{2} x^{2}$

$\left (2 \ln 7 \right )x^{2}$ $\displaystyle \left (2 \ln 7 \right )x^{2}$

$\frac{7}{2}x^{2}$ $\displaystyle \frac{7}{2}x^{2}$

$\frac{ \ln 7}{2}x^{2}$ $\displaystyle \frac{ \ln 7}{2}x^{2}$

$\frac{e^{7}}{2}x^{2}$ $\displaystyle \frac{e^{7}}{2}x^{2}$

Correct answer:

$\frac{\left ( \ln 7 \right )^{2}}{2} x^{2}$ $\displaystyle \frac{\left ( \ln 7 \right )^{2}}{2} x^{2}$

Explanation:

The $x^{2}$ $\displaystyle x^{2}$ term of a Maclaurin series expansion has coefficient

$\frac{f ' ' (0)}{2!} = \frac{f ' ' (0)}{2}$ $\displaystyle \frac{f ' ' (0)}{2!} = \frac{f ' ' (0)}{2}$.

We can find f''(x) $\displaystyle f''(x)$ by differentiating twice in succession:

$f (x) = 7^{x}$ $\displaystyle f (x) = 7^{x}$

$f ' (x) = \frac{\mathrm{d} }{\mathrm{d} x} \left ( 7^{x} \right )= \ln 7 \cdot 7^{x}$ $\displaystyle f ' (x) = \frac{\mathrm{d} }{\mathrm{d} x} \left ( 7^{x} \right )= \ln 7 \cdot 7^{x}$

$f ' '(x) = \frac{\mathrm{d} }{\mathrm{d} x} \left (\ln 7 \cdot 7^{x} \right )$ $\displaystyle f ' '(x) = \frac{\mathrm{d} }{\mathrm{d} x} \left (\ln 7 \cdot 7^{x} \right )$

$f ' '(x)= \ln 7\cdot \frac{\mathrm{d} }{\mathrm{d} x} \left ( 7^{x} \right )$ $\displaystyle f ' '(x)= \ln 7\cdot \frac{\mathrm{d} }{\mathrm{d} x} \left ( 7^{x} \right )$

$f ' '(x)= \ln 7\cdot \ln 7\cdot 7^{x}$ $\displaystyle f ' '(x)= \ln 7\cdot \ln 7\cdot 7^{x}$

$f ' '(x)=\left ( \ln 7 \right )^{2} \cdot 7^{x}$ $\displaystyle f ' '(x)=\left ( \ln 7 \right )^{2} \cdot 7^{x}$

$f ' '(0)=\left ( \ln 7 \right )^{2} \cdot 7^{0} = \left ( \ln 7 \right )^{2} \cdot 1 = \left ( \ln 7 \right )^{2}$ $\displaystyle f ' '(0)=\left ( \ln 7 \right )^{2} \cdot 7^{0} = \left ( \ln 7 \right )^{2} \cdot 1 = \left ( \ln 7 \right )^{2}$

The coefficient we want is

$\frac{f ' ' (0)}{2} =\frac{\left ( \ln 7 \right )^{2}}{2}$ $\displaystyle \frac{f ' ' (0)}{2} =\frac{\left ( \ln 7 \right )^{2}}{2}$,

so the corresponding term is $\frac{\left ( \ln 7 \right )^{2}}{2} x^{2}$ $\displaystyle \frac{\left ( \ln 7 \right )^{2}}{2} x^{2}$.

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High School Math : Calculus II — Integrals

Study concepts, example questions & explanations for High School Math

All High School Math Resources

Example Questions

Example Question #1 : Understanding Maclaurin Series

Example Question #2 : Understanding Maclaurin Series

Example Question #2251 : High School Math

All High School Math Resources

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