The curve with equation y = h(x) passes through the point (4, 19). Given that h'(x) = 15xsqrtx-40/sqrtx, find h(x)?

1 Answer
Mar 4, 2018

since h'(x) = 15x^(3/2) - 40x^(-1/2)

the antiderivative or indefinite integral of h'(x) will be
h(x)
therefore
int 15x^(3/2) - 40x^(-1/2) dx

this can be integrated easily by using the inverse of the Power rule

and it comes out as

2/5 * 15x^(5/3) - 80x^(1/2) + C
=6x^(5/2) - 80x^(1/2) + C

to find C, we can use the fact that h(x) passes through (4,19)
so h(4) = 19

therefore
6*4^(5/2) - 80 * 4^(1/2) + C = 19

6 * 32 - 80 * 2 + C = 19

192 -160 + C = 19

C = 19 -32

therefore

C = -13

therefore, the entire function h(x) is

h(x) = 6x^(5/2) - 80x^(1/2) - 13

and you can factorise it more if you want