[tex] \frac{2(3 - x)^{ \frac{5}{2} } }{5} - \frac{2(3 - x)^{ \frac{3}{2} } }{3} + C[/tex]is the answer
Step-by-step explanation:-
Given:
[tex]{\int {(x-2)\sqrt{3-x}} \ dx }[/tex]
Differentiate on both sides,
[tex]{\int {(x-2)\sqrt{3-x}} \ dx },u = 3 - x[/tex]
Isolate and substitute back,
[tex]{\int {(x-2)\sqrt{3-x}} \ \times ( - 1) du },u = 3 - x[/tex]
Substitute back,
[tex]{\int ( - ( - u + 3 - 2) \sqrt{u})du }[/tex]
Applying property of integral ∫ kf(x)dx = k ∫ f(x)dxdx,
[tex] - ∫ ( - u + 3 - 2) \sqrt{u} \: du,u = 3 - x[/tex]
Combining like terms,
[tex]- ∫ ( - u +1) \sqrt{u} \: du,u = 3 - x[/tex]
Now applying distributive property,
[tex]- ∫ (( - u +1) \sqrt{u}) \: du,u = 3 - x[/tex]
[tex] = > - ∫ ( - \sqrt{u} \times u + \sqrt{u} )\: du,u = 3 - x[/tex]
Converting to exponential form,
[tex]- ∫ ( - u^{ \frac{1}{2} } \times u + u^{ \frac{1}{2} } )\: du,u = 3 - x[/tex]
Multiplying the first two monomuals after integral, we get,
[tex]- ∫ ( - u^{ \frac{3}{2} } + u^{ \frac{1}{2} } )\: du,u = 3 - x[/tex]
Now applying the prperty of ∫ f(x) + g(x)dx = ∫ f(x)dx + ∫ g(x)dx,
[tex] - ( - ∫u^{ \frac{3}{2} } \: du + ∫u^{ \frac{1}{2} } \: du),u = 3 - x[/tex]
Now integrate the power rule,
[tex] - ( - \frac{u^{ \frac{5}{2} }}{ \frac{5}{2} } + \frac{u^{ \frac{3}{2} }}{ \frac{3}{2} } ),u = 3 - x[/tex]
Divide the fractions by multiplying its reciprocals,
[tex] - ( - u^{ \frac{5}{2} } \times \frac{2}{5} + u^{ \frac{3}{2} } \times \frac{2}{3}),u = 3 - x[/tex]
Now write as single fractions,
[tex] - ( - \frac{2u^{ \frac{5}{2} } }{5} + \frac{2u^{ \frac{3}{2} } }{3})[/tex]
Removing the parentheses,
[tex] \frac{2u^{ \frac{5}{2} } }{5} - \frac{2u^{ \frac{3}{2} } }{3},u = 3 - x[/tex]
Substitute back,
[tex] \frac{2(3 - x)^{ \frac{5}{2} } }{5} - \frac{2(3 - x)^{ \frac{3}{2} } }{3} [/tex]
Now adding the constant of integration C∈R,
[tex] \frac{2(3 - x)^{ \frac{5}{2} } }{5} - \frac{2(3 - x)^{ \frac{3}{2} } }{3} + C[/tex]
Hence, the answer.
