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3 years ago

11

Sal rides his bike 24 miles on Sunday that is three times as many miles as he rode his bike on Saturday what is the total number

of miles Sal has ridden his bike on Saturday and Sunday

1 answer:

PtichkaEL [24]3 years ago

5 0

Answer:

32 miles

Step-by-step explanation:

Do 24/3 which =8 and then +24 to it

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Evaluate the following integral (Calculus 2) Please show step by step explanation!

Nuetrik [128]

Answer:

$4\ln \left| \dfrac{1}{3}\sqrt{9+(\ln x)^2} + \dfrac{1}{3}\ln x \right|+\text{C}$

Step-by-step explanation:

<u>Fundamental Theorem of Calculus</u>

$\displaystyle \int \text{f}(x)\:\text{d}x=\text{F}(x)+\text{C} \iff \text{f}(x)=\dfrac{\text{d}}{\text{d}x}(\text{F}(x))$

If differentiating takes you from one function to another, then integrating the second function will take you back to the first with a constant of integration.

Given indefinite integral:

$\displaystyle \int \dfrac{4}{x\sqrt{9+(\ln(x))^2}}\:\:\text{d}x$

Rewrite 9 as 3²:

$\implies \displaystyle \int \dfrac{4}{x\sqrt{3^2+(\ln(x))^2}}\:\:\text{d}x$

<u>Integration by substitution</u>

$\boxed{\textsf{For }\sqrt{a^2+x^2} \textsf{ use the substitution }x=a \tan\theta}$

$\textsf{Let } \ln x=3 \tan \theta$

$\begin{aligned}\implies \sqrt{3^2+(\ln x)^2} & =\sqrt{3^2+(3 \tan\theta)^2}\\ & = \sqrt{9+9\tan^2 \theta}\\ & = \sqrt{9(1+\tan^2 \theta)}\\ & = \sqrt{9\sec^2 \theta}\\ & = 3 \sec\theta\end{aligned}$

Find the derivative of ln x and rewrite it so that dx is on its own:

$\implies \ln x=3 \tan \theta$

$\implies \dfrac{1}{x}\dfrac{\text{d}x}{\text{d}\theta}=3 \sec^2\theta$

$\implies \text{d}x=3x \sec^2\theta\:\:\text{d}\theta$

<u>Substitute</u> everything into the original integral:

$\begin{aligned} \implies \displaystyle \int \dfrac{4}{x\sqrt{9+(\ln(x))^2}}\:\:\text{d}x & = \int \dfrac{4}{3x \sec \theta} \cdot 3x \sec^2\theta\:\:\text{d}\theta\\\\ & = \int 4 \sec \theta \:\: \text{d}\theta\end{aligned}$

Take out the constant:

$\implies \displaystyle 4 \int \sec \theta\:\:\text{d}\theta$

$\boxed{\begin{minipage}{7 cm}\underline{Integrating $\sec kx$}\\\\$\displaystyle \int \sec kx\:\text{d}x=\dfrac{1}{k} \ln \left| \sec kx + \tan kx \right|\:\:(+\text{C})$\end{minipage}}$

$\implies 4\ln \left| \sec \theta + \tan \theta \right|+\text{C}$

$\textsf{Substitute back in } \tan\theta=\dfrac{1}{3}\ln x :$

$\implies 4\ln \left| \sec \theta + \dfrac{1}{3}\ln x \right|+\text{C}$

$\textsf{Substitute back in } \sec\theta=\dfrac{1}{3}\sqrt{9+(\ln x)^2}:$

$\implies 4\ln \left| \dfrac{1}{3}\sqrt{9+(\ln x)^2} + \dfrac{1}{3}\ln x \right|+\text{C}$

Learn more about integration by trigonometric substitution here:

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2 years ago

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Answer:

1/20

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Step-by-step explanation:

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