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

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Answer 1 · 2018-03-04T13:08:53

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