1. The four subintervals are [0, 2], [2, 5], [5, 6], and [6, 7]. Their respective right endpoints are 2, 5, 6, and 7. If [tex]C(t)[/tex] denotes the change in sea level [tex]t[/tex] years after 2010, then the total sea level rise over the course of 2010 to 2017 is[tex]\displaystyle\int_0^7C(t)\,\mathrm dt[/tex]approximated by the Riemann sum,[tex]C(2)(2-0)+C(5)(5-2)+C(6)(6-5)+C(7)(7-6)\approx\boxed{20\,\mathrm{mm}}[/tex]2. The sum represents the definite integral[tex]\boxed{\displaystyle\int_1^4\sqrt x\,\mathrm dx}[/tex]That is, we partition the interval [1, 4] into [tex]n[/tex] subintervals, each of width [tex]\dfrac{4-1}n=\dfrac3n[/tex]. Then we sample [tex]n[/tex] points in each subinterval, where [tex]1+\dfrac{3k}n[/tex] is the point used in the [tex]k[/tex]th subinterval, then take its square root.3. The integral is trivial:[tex]\displaystyle\int_1^4\sqrt x\,\mathrm dx=\frac23x^{3/2}\bigg|_{x=1}^{x=4}=\boxed{\frac{14}3}[/tex]4. Using the fundamental properties of the definite integral, we have[tex]\displaystyle\int_1^4f(x)\,\mathrm dx=e^4-e\implies2\int_1^4f(x)\,\mathrm dx=2e^4-2e[/tex][tex]\displaystyle\int_1^4(2f(x)-1)\,\mathrm dx=2e^4-2e-\int_1^4\mathrm dx=\boxed{2e^4-2e-3}[/tex]5. First note that [tex]\sec x[/tex] is undefined at [tex]x=\dfrac\pi2[/tex], so the integral is improper. Recall that [tex](\tan x)'=\sec^2x[/tex]. Then[tex]\displaystyle\int_0^{\pi/2}\sec^2\frac xk\,\mathrm dx=\lim_{t\to\pi/2^-}\int_0^t\sec^2\frac xk\,\mathrm dx[/tex][tex]=\displaystyle\lim_{t\to\pi/2^-}k\tan\frac xk\bigg|_{x=0}^{x=t}[/tex][tex]=\displaystyle k\lim_{t\to\pi/2^-}\tan\frac tk[/tex][tex]=k\tan\dfrac\pi{2k}[/tex]Now,[tex]k\tan\dfrac\pi{2k}=k\implies\tan\dfrac\pi{2k}=1[/tex][tex]\implies\dfrac\pi{2k}=\dfrac\pi4+n\pi[/tex][tex]\implies k=\dfrac2{1+4n}[/tex]where [tex]n[/tex] is any integer.