It is given that $y=x^2\sqrt{2x-5}$, whose natural real domain is $x\ge h$. Show that

Question

It is given that $y=x^2\sqrt{2x-5}$, whose natural real domain is $x\ge h$. Show that $\frac{dy}{dx}=\frac{kx(x-2)}{\sqrt{2x-5}}$, where $k$ is an integer and determine the value of $h$ and of $k$.[4] Hence evaluate $\displaystyle \int_{3}^{7}\left\{\frac{10x(x-2)}{\sqrt{2x-5}}+6\right\}\,dx$.[4] The derivative formula in part (a) applies for $x>h$; the function is defined at $x=h$, but its finite derivative is not.

Accepted Answer

$h=\frac{5}{2},\ k=5$ $(b)\ \displaystyle\int_3^7\left\{\frac{10x(x-2)}{\sqrt{2x-5}}+6\right\}\,dx=300$

Integration - Definite Integration as the Reverse of Differentiation

Integration - Definite Integration as the Reverse of Differentiation

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