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Evaluating Integral
- Thread starter q_fruit
- Start date
let u = 3-2s
du= -2ds
-1/2du = ds
(sorry im not familiar with TeX so i'll just use * to represent the integral sign)
*(3-2s)^1/2ds
= *u^1/2 x 1/2du
= -1/2 * u^1/2du
= -1/2 x (2/3)(u^3/2) + C
= -1/2 x (2/3)([3-2s]^3/2) + C
= -(2/6)(3-2s)^3/2 + C
du= -2ds
-1/2du = ds
(sorry im not familiar with TeX so i'll just use * to represent the integral sign)
*(3-2s)^1/2ds
= *u^1/2 x 1/2du
= -1/2 * u^1/2du
= -1/2 x (2/3)(u^3/2) + C
= -1/2 x (2/3)([3-2s]^3/2) + C
= -(2/6)(3-2s)^3/2 + C
Looks great, q_fruit! Perhaps they just wrote it differently?
\(\displaystyle \L -\frac{1}{3}(3 - 2s)^{\frac{3}{2}} \, + \, C\)
or
\(\displaystyle \L -\frac{1}{3}\sqrt{(3 - 2s)^3 \, } \, + \, C\)
or
\(\displaystyle \L -\frac{(3 - 2s)^{\frac{3}{2}}}{3} \, + \, C\)
or
\(\displaystyle \L -\frac{\sqrt{(3 - 2s)^3 \, }}{3} \, + \, C\)
That's me out of ideas.
\(\displaystyle \L -\frac{1}{3}(3 - 2s)^{\frac{3}{2}} \, + \, C\)
or
\(\displaystyle \L -\frac{1}{3}\sqrt{(3 - 2s)^3 \, } \, + \, C\)
or
\(\displaystyle \L -\frac{(3 - 2s)^{\frac{3}{2}}}{3} \, + \, C\)
or
\(\displaystyle \L -\frac{\sqrt{(3 - 2s)^3 \, }}{3} \, + \, C\)
That's me out of ideas.