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$$\int \left(4-x(16-x^2)^{1/2}\right)\,dx $$

I learned today I could use U-substitution to before integrating, which makes it easier to integrate.

So I can make $U=16-x^2,\quad \dfrac{du}{dx} = -2x^2$

And I've no idea how I should proceed next.

My teacher only showed me an easy example:

$$x(x+3)^{1/2} \implies U=x+3,\;\; \frac{du}{dx} = 1$$

$$\text{Substitution:}\quad x(x+3)^{1/2} \quad = \quad(U-3)U^{1/2}$$

But I can't figure out on my own how to proceed with the integral on top.

amWhy
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user70994
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  • Notation: when making an $u$-substitution use the same variable: in the definition $u=16-x^2$ and in the derivative $\frac{du}{dx}=-2x^2$ (or $U=16-x^2$, $\frac{dU}{dx}=-2x^2$), instead of $U=16-x^2$, $\frac{du}{dx}=-2x^2$. In Mathematics lower case letters are normally used for variables. – Américo Tavares Apr 05 '13 at 00:48

3 Answers3

3

You chose the correct $$U:\;\;\color{red}{\bf U = 16 - x^2}\implies dU = -2x\,dx \implies \color{blue}{\bf x\,dx = -\frac 12\, dU}\tag{1}$$

Substituting terms of $(2)$ into the equivalent substitutions shown in $(1)$ gives us $(3)$: $$\int \left(4-x(16-x^2)^{1/2}\right)\;dx = \int 4 dx - \int \color{red}{\bf \left(16-x^2\right)}^{1/2}\,\color{blue}{\bf x\,dx}\tag{2}$$ $$= \int 4 \;dx - \left(\color{blue}{\bf -\frac 12}\right)\int \color{red}{\bf U}^{1/2} \,\color{blue}{\bf dU}\tag{3}$$

That is, we compute the integrals: $$\int 4 \,dx + \left(\frac 12\right)\int U^{1/2} \, dU\quad = \quad 4x + \left(\frac 12\cdot \frac 23 \right) U^{3/2} + C$$ $$ = 4x+ \left(\frac 13\right)\left(16-x^2\right)^{3/2} + C\tag{4}$$

$(4)$ When using U-substitution, don't forget to "back-substitute"!

amWhy
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  • I hope this helps you to visualize what's going on with u-substitution. You can also search the Math.SE site using key words "integral" and "u-substitution" and you'll find lots of examples where it's been used. – amWhy Apr 04 '13 at 22:42
2

Since you have $$U=16-x^2$$ and $$dU=-2xdx$$ (Notice your mistake)

Continue what you are doing.

$$\int4dx+\frac{1}{2}\int U^{1/2}dU$$

1

As you said, define $u:=16-x^2$. The only thing wrong is the derivative, $\frac{du}{dx} = -2x \ dx$ Substituting will cause the $x$ to fall out.

Henry Swanson
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