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4 years ago

Round 7.92 + 5.37 to estimate

Mathematics

1 answer:

maria [59]4 years ago

8 0

Answer:

13.4

Step-by-step explanation:

You can round 7.92 to 8 and 5.37 to 5.4

You then add 8 and 5.4 to get 13.4

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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:

Fundamental Theorem of Calculus

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

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:

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

Integration by substitution

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

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

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

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

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

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

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

Substitute 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})$\end{minipage}}$