Solve the integral and show work. 3 is on the top of the integral and 0 is on the bottom of the integral

Question

₀³∫x√9-x² dx

Answer 1

integral of xsqr(9-x)dx can be found in integration tables

= -(1/3)(9-x^2)^(3/2)

ignore the constant for a definite integral

evaluated from 0 to 3

=0- -(1/3)(9^(3/2)) =(1/3)sqr(9^3) = (1/3)(sqr(81(9)) = (1/3)(27)/3 = 9

= 9

Answer 2

Let u = 9 - x² then du/dx = -2x or dx = du/(-2x) so rewriting the integral:

₀∫³(x(u^1/2)/(-2x))du

(-1/2)₀∫³(u^1/2)du

(-1/2(2/3)(u^3/2) evaluated between 0 and 3

(-1/3)(u^3/2) evaluated between 0 and 3

Back-substituting 9 - x² = u:

(-1/3)(9 - x²)^3/2 evaluated between 0 and 3

(-1/3)(9 - 3²)^3/2 - (-1/3)(9 - 0²)^3/2

(-1/3)(0)^3/2 - (-1/3)(9)^3/2

0 + (1/3)(27)

9

Answer 3

Evaluate ∫₀³ x/√(9-x²)dx

Use u-substitution.

Let u= 9-x², du=-2xdx

dx= -du/(2x)

Converting the integration limits to u-space, when x=0, u=9, and when x=3, u=0. So the integral becomes

∫₉⁰ x/(-2)x√udu

= -∫₉⁰ 1/2√udu

= -√u |₉⁰

= 3