Respuesta :

Answer:

A) [tex]\int\frac{2x^2}{3}dx=\frac{2x^3}{9}+C[/tex]

B) [tex]\int(5-x)^{23}dx=-\frac{(5-x)^{24}}{24}+C[/tex]

Step-by-step explanation:

A)

So we have the integral:

[tex]\int\frac{2x^2}{3}dx[/tex]

First, remove the constant multiple:

[tex]=\frac{2}{3}\int x^2\dx[/tex]

Use the power rule, where:

[tex]\int x^ndx=\frac{x^{n+1}}{x+1}[/tex]

Therefore:

[tex]\frac{2}{3}\int x^2\dx\\=\frac{2}{3}(\frac{x^{2+1}}{2+1})[/tex]

Simplify:

[tex]=\frac{2}{3}(\frac{x^{3}}{3})[/tex]

And multiply:

[tex]=\frac{2x^3}{9}[/tex]

And, finally, plus C:

[tex]=\frac{2x^3}{9}+C[/tex]

B)

We have the integral:

[tex]\int(5-x)^{23}dx[/tex]

To solve, we can use u-substitute.

Let u equal 5-x. Then:

[tex]u=5-x\\du=-1dx[/tex]

So:

[tex]\int(5-x)^{23}dx\\=\int-u^{23}du[/tex]

Move the negative outside:

[tex]=-\int u^{23}du[/tex]

Power rule:

[tex]=-(\frac{u^{23+1}}{23+1})[/tex]

Add:

[tex]=-(\frac{u^{24}}{24})[/tex]

Substitute back 5-x:

[tex]=-(\frac{(5-x)^{24}}{24})[/tex]

Constant of integration:

[tex]=-\frac{(5-x)^{24}}{24}+C[/tex]

And we're done!

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