How do you find the definite integral: #int_0^4 4 sqrtx dx#?

Question

1907 views around the world

You can reuse this answer
Creative Commons License

Answer 1 · 2017-07-02T21:56:00

Given #int_0^4 4 sqrtx dx = ?#

Write the square root as the #1/2# power:

#int_0^4 4 sqrtx dx =int_0^4 4 x^(1/2) dx#

Use the power rule, #intx^n dx = 1/(n+1)x^(n+1)#:

#int_0^4 4 sqrtx dx =4(2/3) (x^(3/2)]_0^4#

#int_0^4 4 sqrtx dx =8/3 (4^(3/2)-0^(3/2))#

#int_0^4 4 sqrtx dx =64/3#