How do you find the definite integral of #int x(x^(2/3))dx # from #[0,1]#?

Question

Isabella Ackerman · Accepted Answer

#int_0^1 x(x^(2/3))d x=3/8# #int_0^1 x(x^(2/3))d x=int_0^1 x root(3)(x^2)d x# #int_0^1 x(x^(2/3))d x=int_0^1 root(3)(x^3*x^2)d x# #int_0^1 x(x^(2/3))d x=int_0^1 x^(5/3) d x# #int_0^1 x(x^(2/3))d x=|3/8*x^(8/3)|_0^1# #int_0^1 x(x^(2/3))d x=3/8*1^(8/3)-3/8*0^(8/3)# #int_0^1 x(x^(2/3))d x=3/8#

Isaiah Allen · Answer

undefined To find the definite integral of $ \int_0^1 x(x^{2/3}) \, dx $, you can follow these steps:

1. Use the power rule for integration to integrate $ x(x^{2/3}) $.
   $$ \int x(x^{2/3}) \, dx = \int x^{5/3} \, dx $$

2. Apply the power rule for integration, which states that $ \int x^n \, dx = \frac{x^{n+1}}{n+1} + C $.
   $$ \int x^{5/3} \, dx = \frac{3}{8}x^{8/3} + C $$

3. Evaluate the definite integral by substituting the upper and lower limits of integration (1 and 0, respectively) into the antiderivative and subtracting the results.
   $$ \int_0^1 x(x^{2/3}) \, dx = \left[\frac{3}{8}x^{8/3}\right]_0^1 $$
   $$ = \left(\frac{3}{8}(1)^{8/3}\right) - \left(\frac{3}{8}(0)^{8/3}\right) $$
   $$ = \frac{3}{8} - 0 $$
   $$ = \frac{3}{8} $$

Therefore, the definite integral of $ \int_0^1 x(x^{2/3}) \, dx $ is $ \frac{3}{8} $.