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How to evaluate this definite integrand: $$\int_{-b}^b \frac{\pi}{a^4}\left(y^2-b^2\right)^4 \,\mathrm{d}y$$

Michael Hardy
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Ali khan
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    Is this the integral $\int_{-b}^{b} (\frac{\pi}{a^4})(y^2 - b^2)^4 dy$? I ask, because I am having a little difficulty interpreting what you wrote. – Mack Dec 07 '13 at 20:33
  • How about you apply the binomial formula twice and then you are left with integrands of the form $y^j$ with j being natural constants. –  Dec 07 '13 at 20:44
  • @Nabla Why would you have to apply it twice? Wouldn't once suffice? – Mack Dec 07 '13 at 20:46
  • Since the question looks kind of simple I was referring to the special case of the binomial formula with exponent 2. This is what we used to call THE binomial formula in school. –  Dec 07 '13 at 20:52
  • yes Mack you interpreted rightly. – Ali khan Dec 07 '13 at 21:49
  • Nabla,can you solve it please? – Ali khan Dec 07 '13 at 22:10
  • can anyone explain the solution more clearly? – Ali khan Dec 08 '13 at 08:32

2 Answers2

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Let $y=b\cos(t)$. Hence, we have $$\int_{-b}^{b} \left(y^2-b^2\right)^4 dy = \int_{0}^{\pi} b^9 \sin^9(t)dt = 2b^9\int_0^{\pi/2} \sin^9(t)dt$$ From here, we have $$\int_0^{\pi/2} \sin^9(t)dt = \dfrac89 \cdot \dfrac67 \cdot \dfrac45 \cdot \dfrac23$$

Community
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All right, I suppose I'll take a stab at this.

Notice, that for the function we are integrating, we are integrating between the intercepts of the function. So, if we call the function for which we are integrating $g(y)$, then $g(y) = \displaystyle \frac{\pi}{a^4} (y^2 - b^2)$, and if we were to evaluate the function at $-b$ and $b$, it would yield $g(-b) = 0$ and $g(b) = 0$.

Now, let us examine the graph of the function for which we are integrating; but to make things a little more concrete, I am going to allow $b = 2$, but the analysis is the same for $b$. Here is a link to the graph of $g(y) = (y^2 - 2^2)$,

Notice the area over which you are integrating is symmetric about the vertical axis, that is, the area between $-2$ and $0$ is the same as between $0$ and $2$. Going back to our original function, involving $b$, we can slightly simplify the integral by integrating between $0$ an $b$, and multiplying that integral by $2$:

$$ \frac{2 \pi}{a^4} \int_{0}^{b} (y^2 - b^2)^2 \mathrm{d}y.$$

At this point, the most simple alternative would be to foil the expression $(y^2 - b^2)^2$, and then integrate; although, you could use the binomial theorem, as suggested by Nabla.

Is this sufficiently clear?

Bennett Gardiner
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Mack
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