Integral

[CatchThis2]

I broke the equation down to:

Bring the 7 out in front of the integral sign and behind it I have x^-1/2 minus 5 integral sign x^2/3 dx

Now I have added n+1 and got 7x^1/2 -5/2x^5/3

 

[galactus]

Not quite. You added the 1 to the exponent, but did you divide that into the coefficient?.

\(\displaystyle \int\frac{7}{x^{\frac{1}{3}}}dx=7\int x^{\frac{-1}{3}}dx\)

Adding one to the exponent gives 2/3. But, 7 divided by 2/3 is 21/2.

So, we get \(\displaystyle \frac{21}{2}x^{\frac{2}{3}}\)

Try fixing the other one. You have the exponent correct, but the coefficient is incorrect.

Add 1 to the exponent first, then divide that into the existing coefficient to get the new one.

It's the opposite of differentiating.

 

[CatchThis2]

So add one to (-1/3)and divide by (-1/3)?

 

[galactus]

No. For the first one, add one to -1/3 and divide into 7.

7/(2/3)=21/2.


For the second one, add one to 2/3 and divide that into 5(the existing coefficient).


1+2/3=5/3. 5/(5/3)=3.

So, you get \(\displaystyle 3x^{\frac{5}{3}}\)

See now?.

 

[CatchThis2]

I figured it out from the step you left me at. Thank you for your help!