Evaluating Integrals

[Silvanoshei]

Could you please check my work?

\(\displaystyle ∫\sqrt{3-2s}ds\)

\(\displaystyle u=3-2s\)

\(\displaystyle \frac{d}{dx}3-2s=-2\)\(\displaystyle ds=\frac{du}{-2}\)

\(\displaystyle ∫u^{\frac{1}{2}}\frac{du}{-2}\)

\(\displaystyle -\frac{1}{2}∫u^{\frac{1}{2}}du\)

\(\displaystyle -\frac{1}{2}\frac{u^{\frac{3}{2}}}{\frac{3}{2}}+c\)

\(\displaystyle -\frac{1}{2}u^{\frac{3}{2}}*\frac{2}{3}+c\)

\(\displaystyle -\frac{1}{3}(3-2s)^{\frac{3}{2}}+c\) ???

 

[Deleted member 4993]

Could you please check my work?

\(\displaystyle ∫\sqrt{3-2s}ds\)

\(\displaystyle u=3-2s\)

\(\displaystyle \frac{d}{dx}3-2s=-2\)\(\displaystyle ds=\frac{du}{-2}\)

\(\displaystyle ∫u^{\frac{1}{2}}\frac{du}{-2}\)

\(\displaystyle -\frac{1}{2}∫u^{\frac{1}{2}}du\)

\(\displaystyle -\frac{1}{2}\frac{u^{\frac{3}{2}}}{\frac{3}{2}}+c\)

\(\displaystyle -\frac{1}{2}u^{\frac{3}{2}}*\frac{2}{3}+c\)

\(\displaystyle -\frac{1}{3}(3-2s)^{\frac{3}{2}}+c\) ???

You can check your work by differentiating the anti-derivative.

If you get beck the original function after differentiation, you are probably correct.

 

[lookagain]

Could you please check my work?

\(\displaystyle ∫\sqrt{3-2s}ds\)

\(\displaystyle u=3-2s\)

\(\displaystyle \frac{d}{dx}3-2s=-2\)\(\displaystyle ds=\frac{du}{-2}\)\(\displaystyle \ \ <--- Make \ \ this \ \ \frac{d}{ds}(3-2s)=-2, \ \ and \ \ then \ \ ds = \frac{du}{-2}.\)

Instead, I would type these:

\(\displaystyle u=3-2s\)

\(\displaystyle du \ = \ -2ds\)

\(\displaystyle \frac{du}{-2} \ = ds \ \ or \)

\(\displaystyle ds \ = \ \frac{du}{-2} \ = \ (\frac{-1}{2})du\)


.