what is the integral of x(x^2+1)^(1/3) with upper limit 0 and lower limit -(7^(1/2))

Compute the definite integral:

integral_(-sqrt(7))^0 x (x^2+1)^(1/3) dx

For the integrand x (x^2+1)^(1/3), substitute u = x^2+1 and du = 2 x dx.

This gives a new lower bound u = (-sqrt(7))^2+1 = 8 and upper bound u = 1+0^2 = 1:

= 1/2 integral_8^1 u^(1/3) du

Switch the order of the integration bounds of u^(1/3) so that the upper bound is larger. Multiply the integrand by -1:

= -1/2 integral_1^8 u^(1/3) du

Apply the fundamental theorem of calculus.

The antiderivative of u^(1/3) is (3 u^(4/3))/4:

= (-(3 u^(4/3))/8)|_1^8

Evaluate the antiderivative at the limits and subtract.

(-3/8 u u^(1/3))|_1^8 = (-3/8 8 8^(1/3))-(-(3 1^(1/3))/8) = -45/8:

Answer: |= -45/8