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3 years ago

Write an equation in slope-intercept form for the line with slope -1/4 and y-intercept 9

Mathematics

2 answers:

Nat2105 [25]3 years ago

4 0

$y=- \frac{1}{4} x+9$

vichka [17]3 years ago

3 0

The answer is y=-1/4x+9

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3 years ago

Use integration by parts to find the integrals in Exercise. x^3 ln x dx.

34kurt

Answer:

$\frac{\text{ln}(x)x^4}{4}-\frac{x^4}{16}+C$ .

Step-by-step explanation:

We have been given an indefinite integral $\int \:x^3\:ln\:x\:dx$ . We are asked to find the value of the integral using integration by parts.

$\int\: u\text{dv}=uv-\int\: v\text{du}$

Let $u=\text{ln}(x)$ , $v'=x^3$ .

Now, we will find du and v as shown below:

$\frac{du}{dx}=\frac{d}{dx}(\text{ln}(x))$

$\frac{du}{dx}=\frac{1}{x}$

$du=\frac{1}{x}dx$

$v=\frac{x^{3+1}}{3+1}=\frac{x^{4}}{4}$

Upon substituting our values in integration by parts formula, we will get:

$\int \:x^3\:\text{ln}\:(x)\:dx=\text{ln}(x)*\frac{x^4}{4}-\int\: \frac{x^4}{4}*\frac{1}{x}dx$

$\int \:x^3\:\text{ln}\:(x)\:dx=\frac{\text{ln}(x)x^4}{4}-\int\: \frac{x^3}{4}dx$

$\int \:x^3\:\text{ln}\:(x)\:dx=\frac{\text{ln}(x)x^4}{4}-\frac{1}{4}\int\: x^3dx$

$\int \:x^3\:\text{ln}\:(x)\:dx=\frac{\text{ln}(x)x^4}{4}-\frac{1}{4}*\frac{x^{3+1}}{3+1}+C$

$\int \:x^3\:\text{ln}\:(x)\:dx=\frac{\text{ln}(x)x^4}{4}-\frac{1}{4}*\frac{x^4}{4}+C$

$\int \:x^3\:\text{ln}\:(x)\:dx=\frac{\text{ln}(x)x^4}{4}-\frac{x^4}{16}+C$

Therefore, our required integral would be $\frac{\text{ln}(x)x^4}{4}-\frac{x^4}{16}+C$ .

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3 years ago