Finding the area under the first three curves

Question

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)

Answer 1

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 $

Answer 2

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 $f(x)=\dfrac{1+\sin(2\pi x)}{1+x}$