Question

Evaluate the integral. ?

Answer 1

3

Explanation:

Using u-substitution and the rules of anti derivatives.
Let `u=27+2x`
`du=2dx`; `(du)/2=dx` or `du*1/2=dx`
Go back to the original formula and plug that in
`int_0^49(1/root(3)(u^2)du*1/2)`
Using properties of integrals, I can put constants in front of the integral.
`1/2int_0^49(1/root(3)(u^2))du`
Using properties of roots and exponents:
`a^(x/y)=root(y)(a^x)`
and `1/x=x^-1`
Now convert again to
`1/2int_0^49(u^(-2/3))du`
Apply reverse power rule to u and then get
`1/2[3root(3)(u)]_0^49`
Replace u with original function and remember
`int_a^bf(x)dx=F(b)-F(a)`
so then
`1/2[3root(3)(27+2(49))-3root(3)(27+2(0))]` and you should hopefully get 3.
Have fun!