Solve the integral and show work. The 4 is the bottom part of the integral while the 9 is the top part of the integral

Question

₄⁹∫x-3/√x dx

Answer 1

integral of x -3/x^.5

= integral of x -3x^-.5

= x^2/2 -3x^.5/.5

= x^2/2 -6x^.5

evaluated from 4 to 9

= 81/2 -6(3) -(16/2 -6(2))

=81/2-18-8+12

=40.5-14

=26.5

= 26 1/2

UNLESS you really meant the integral of (x-3)/x^.5

then

= integral of x^.5 -3x^-.5

= x^1.5/1.5 -3x^.5/.5

= (2/3)x^1.5 - 6x^.5

evaluated from 4 to 9

= (2/3)9sqr9 -6sqr9 - ((2/3)4sqr4 - 6sqr4)

= 18-18 - (2/3)(8) + 12

= 12- 16/3

= 20/3

= 6 2/3

as written

the integral of x-3/sqrx

= the integral of x - (3/sqrx)

and the answer is 26 1/2

but

occasionally parentheses accidentally get left out

so it's possible the problem was supposed to read

the integral of (x-3)/sqrx

then

the answer is 6 2/3

Answer 2

∫₄⁹x-3/√x dx = ∫₄⁹ xdx - ∫₄⁹ 3/√x dx

                     = ½ x²|₄⁹ - 6√x |₄⁹

                     = ½(9²-4²) – 6(√9 - √4)

                     = 65/2 – 6

                     = 53/2

                     = 26.5

Answer 3

∫_{(from 4 to 9)} [(x - 3) / √x]dx = ∫_{(from 4 to 9)}[(x / √x ) - (3 / √x)]dx = ∫_{(from 4 to 9)} [√x - 3x^-1/2]dx

= [(2/3)x^3/2 - 6√x]_{(from 4 to 9)} = [18 - 18] - [16/3 - 12] = 20/3

Answer 4

Please see video. Hope this helps!

Ibrahim