# Tricky integral

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1. Jan 14, 2016

### PhysicsBoyMan

1. The problem statement, all variables and given/known data
Evaluate the integral to find the area.
2. Relevant equations
3. The attempt at a solution

So I know how to find an anti-derivative for the most part. Here it's tricky because my equation has an exponent AKA square root. I tried to use the chain rule with anti-differentiation. My result was 0. I'm not sure how to deal with that exponent AKA square root.

2. Jan 14, 2016

### blue_leaf77

Never heard of that. It doesn't seem to work, does it? Try to use the substitution method with the help of trigonometric identity, remember that $a^2\sin^2\theta + a^2\cos^2\theta = a^2$.
Note: exponent is not the same as square root.

3. Jan 15, 2016

### HallsofIvy

Staff Emeritus
The "chain rule with integration" is "substitution"- but you are doing it wrong.
If you had $\int (ax- b)^n dx$ then you could let u= ax- b so that du= adx and dx= du/a. Replacing ax- b by u and dx by du/a, the integral becomes $\int u^n (du/a)$ and because a is a constant we can take it out and get $\frac{1}{a}\int u^n du[/tex]. But if we have [itex]\int (cos(x)- b)^n dx$ and try to do the same thing, we run into a problem. Letting u= cos(x)- b, we have du= -sin(x)dx so that dx= -du/cos(x). Putting those into the integral, $\int u^n (du/cos(x))$. But now we cannot take "cos(x)" outside the integral because it is NOT a constant! And we cannot integrate cos(x) with respect to u.

The "chain rule" with differentiation says that if f(x)= g(u(x)) for some differentiable function u, then df/dx= (dg/du)(du/dx). That is, we multiply together the derivatives of g with respect to u and u with respect to x. But doing the "opposite", substituting in the integral for u(x), $\int g(u(x))dx$ we cannot just do the integration with respect to u of g and then multiply by the integral of g. We have to have something line $\int g(u(x))u'(x)dx$. u'dx= du so the integral becomes $\int g(u)du$. But the derivative, u', has to already be in the integral. We cannot just multiply as we can with differentiation.

4. Jan 15, 2016

### Staff: Mentor

You have errors on every line of the work you have done. Here is your work, reproduced using LaTeX, with comments.
$\int_0^{\sqrt{19}} \sqrt{19 - h^2} dh$
$= (19 - h^2)^{1/2}$ No. All you have done is to remove the integration symbol.
$= \frac{2(19 - h^2)^{1/2}}{3} \cdot (19h - \left. \frac{h^3}{3}) \right|_0^{\sqrt{19}}$ See comment below.
$= 0$ Not a reasonable answer. The integral represents the area of a quarter-circle, whose area is not zero.

From above:
No, this isn't how it works. To get $19h - \frac{h^3} 3$ you integrated $19 - h^2$ but ignored the fact that $19 - h^2$ is raised to the 1/2 power. You can verify that your integration is incorrect by differentiating your antiderivative. For example, $\int x^3 dx = \frac {x^4} 4 + C$. If I differentiate $\frac {x^4} 4 + C$, I get $x^3$, the integrand I started with.