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I am asked to calulate the area under the first three curves of the function $$\frac{1 + \sin(2 x \pi)}{1+x}$$ on the interval $[0, 10]$.

I know how to integrate, but I don't know how to find the area under a specific section (like here, three curves)

Math-fun
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Garth Marenghi
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  • First find the three positives roots. Then use any numerical integration method up to calculate the integral from $0$ to the third root. – Math-fun Apr 16 '15 at 08:12
  • I get what you're saying. So... can I find these roots in maxima easily? I saw there's a root finder, but it said it doesn't accept variables. – Garth Marenghi Apr 16 '15 at 09:38
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    Roots are where the numerator becomes zero. This happens when $\sin 2\pi x =-1$. Using the properties of $\sin$ function we find that the first root is $x_1=3/4$, the second is $x_2=7/4$, and $x_3=11/4$. – Math-fun Apr 16 '15 at 12:35

2 Answers2

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Hint: you should find the first three postive roots$(x_0,x_1,x_2)$ and then $$\int_{0}^{x_0}f(x)dx+\int_{x_0}^{x_1}f(x)dx+\int_{x_1}^{x_2}f(x)dx$$ so $f(x)=\dfrac{1+\sin(2\pi x)}{1+x}$

Narasimham
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E.H.E
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There is only one curve for one equation. Sketch the curve. There are no first three curves. You want sum of areas under three humps or arch portions starting and terminating on the positive x-axis.

Find/solve or verify that (double) roots exist for $ x = ( n -\frac 14) $ where the curved arches touch the x-axis. Either you can evaluate areas under arches separately:

$$\int_{\frac{3}{4}}^{\frac{7}{4}}f(x)dx + \int_{\frac 74 }^{\frac{11}{4}}f(x)dx+\int_{{\frac{11}{4}}}^{{\frac{15}{4}}}f(x)dx $$

or even a single integral would suffice for the same purpose:

$$\int_{{\frac{3}{4}}}^{{\frac{15}{4}}}f(x)dx $$

EDIT1

To find location of roots involving negative arcsin: $ 2 \pi x = ( 4 n - 1) \pi/2 $

Narasimham
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  • "There are no first three curves" - No, you're correct, since it is a continuous function. What I meant to say was that if you only look at the interval [0, 10], so starting from x = 0, from that point onward the area of the first three "humps". – Garth Marenghi Apr 16 '15 at 09:50