Question

Calculas help

Answer 1

@LastDayWork @quixoticideals @Destinymasha @FATBOI @ybarrap @Compassionate @ZekeWars

Answer 2

looks like the forth choice

Answer 3

@sourwing how do u know?

Answer 4

@LastDayWork @lululife

Answer 5

@sourwing

Answer 6

If the interval is from 0 to 3 then each partition has size \(\cfrac{3-0}{n}=\cfrac{3}{n}\)
So, we are given
$$
\large{
\lim_{n\to \infty}\cfrac{3}{n}\Sigma_{k=1}^{n}\left (\cfrac{6k}{n}+\cfrac{\sin(6k\pi}{n}\right)\\
}
$$
But 3/n is our interval, lets call it \(\Delta x\), then
$$
\large{
\lim_{\Delta x\to 0}\Delta x\times\Sigma_{k=1}^{n}\left (2k\Delta x+\sin(2k\Delta x\pi\right)
}
$$
Which equals in the limit:
$$
\large
\int_{0}^{3}\left (2x+\sin(2\pi x)\right )~dx
$$

The second one is handled similarly (omitting extraneous steps), and \(\Delta x=\cfrac{2}{n}\)
$$
\large{
\lim_{\Delta x\to 0}\Delta x \times \Sigma_{k=1}^{n}\left (2+(2k\Delta x)^3\right)
=\\
\int_0^2\left (2+(2x)^3 \right )~dx
}
$$

Let me know if you have any questions.