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Question

(a) integrate to find F as a function of x, and (b) demonstrate the Second Fundamental Theorem of Calculus by differentiating the result in part (a). F(x)=4xt3/2dtF(x)=\int_{4}^{x} t^{3 / 2} d t

Solution

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a)

F(x)=4xt3/2 dt=[t5/25/2]4x=25(x5/205/2)=25x5\begin{align*} F(x)=&\int_4^x t^{3/2}\ d t\\ =&\left[\frac{t^{5/2}}{5/2}\right]_4^x\tag{general power and constant rules}\\ =&\frac25\left(x^{5/2}-0^{5/2}\right)\tag{substitute by the limits}\\ =&\frac25\sqrt{x^5}\tag{simplify}\\ \end{align*}

b)

d dxF(x)=d dx(25x5)=25d dx x5/2=2552x3/2=x3/2=f(x)\begin{align*} \dfrac{d\ }{dx}F(x)=&\dfrac{d\ }{dx}\left(\frac25\sqrt{x^5}\right)\tag{from (a)}\\ =&\frac25\dfrac{d\ }{dx}\ x^{5/2}\tag{multiple constant rule}\\ =&\frac25\cdot\frac52x^{3/2}=x^{3/2}=f(x)\tag{power rule}\\ \end{align*}

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