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!
