I'm confused about part B of the following exercise on basic integrals. I am asked to find the area of R2, the blue shaded area.

My method was to find the area of the area of the triangle (0,7), (0,0), (7,0) using $1/2bh$ or integration, then to subtract the area of the parabola between $x = 0$ and $x = 2$ (intersection) and the line.



However the printed solution is quite different and I don't understand why it is correct and my reasoning is not. They seem to have taken the area of the smaller triangle between $x = 2$ (right-most intersection of the line and the parabola) and $x = 7$, then added the area between the parabola, line and positive y-axis.

But doesn't that miss the part of R2 (blue area) between $x=2$ and the y-axis, below the parabola?



Thanks very much for your help!

p.s. Another quick, related question: Am I correct in thinking that in order to find the area under a curve, you only need to split the integral into two sums if the curve goes below the x-axis?

We can bound the partial sums by multiples of themselves:
$$\begin{eqnarray}
S_{2k+1}
&=&
\sum_{n=1}^{2k+1}\frac{1}{n^p}\\
&=&
1+\sum_{i=1}^k\left(\frac{1}{(2i)^p}+\frac{1}{(2i+1)^p}\right)\\
&<&1+\sum_{i=1}^k\frac{2}{(2i)^p}\\
&=&1+2^{1-p}S_k\\
&<&1+2^{1-p}S_{2k+1}\;.
\end{eqnarray}$$
Then solving for $S_{2k+1}$ yields
$$S_{2k+1}<\frac{1}{1-2^{1-p}}\;,$$
and since the sequence of partial sums is monotonically increasing and bounded from above, it converges.

Eight years after I posted this, Zach Teitler pointed out that this proof was already published in $1979$ by Teresa Cohen and William J. Knight in Convergence and Divergence of $\sum_{n=1}^{\infty} 1/n^p$, Mathematics Magazine, $52(3)$, $1979$, p. $178$ (accessible version, DOI link).
