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.

Confused about part b

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?

Confused about this solution

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?

  • $\begingroup$ What do you mean by "the area of the parabola between x=0 and x=7 "? $\int_0^7 (x^2+1) dx$ ? $\endgroup$
    – shamovic
    Mar 28, 2011 at 7:27
  • $\begingroup$ Oops sorry, that should be "between x=0 and x=2". I'll change that now. $\endgroup$
    – Danny King
    Mar 28, 2011 at 7:30
  • $\begingroup$ Or just in case I'm still not being clear, my method was: find the area of the triangle then subtract the yellow region that lies on the +ve x-axis. $\endgroup$
    – Danny King
    Mar 28, 2011 at 7:31
  • $\begingroup$ I think "lies on the +ve x-axis" in my comment above should read "lies in the 1st quadrant" $\endgroup$
    – Danny King
    Mar 28, 2011 at 7:40

2 Answers 2


Your method, conceptually, should work. The part of the yellow region that you are subtracting from the triangle should be $\int_0^2(7-x-(x^2+1))dx$, since it's the area bounded by $y=7-x$ above and $y=x^2+1$ below, from $x=0$ to $x=2$. In their solution, $\int_0^2(x^2+1)dx$ is the portion of the blue region bounded by $y=x^2+1$ above, the $y$-axis below, from $x=0$ to $x=2$.

As to your p.s., it depends on exactly what is meant by "area under a curve." For the geometric, unsigned area between the $x$-axis and the curve, you'd probably want to integrate over intervals where the function has one sign (where the function is only positive or is only negative).


The integral $\int_0^2 (x^2+1) dx$ that they add to the triangle area is exactly the area of that part that you're worrying that they are missing.

  • $\begingroup$ Interesting, thanks! In that case, how did you know this and also, was my method incorrect? (I got a different answer). $\endgroup$
    – Danny King
    Mar 28, 2011 at 7:38
  • $\begingroup$ @Danny: Well, the integral $\int_a^b f(x) dx$ is (if $f\ge 0$) simply the area below the graph $y=f(x)$ between $x=a$ and $x=b$. So the integral $\int_0^2 (x^2+1) dx$ is the area below the parabola $y=x^2+1$. It's not "the area between the parabola, line and positive y-axis". How could it have anything to do with the line, when the formula doesn't take the line's equation $y=7-x$ into account at all? $\endgroup$ Mar 28, 2011 at 9:32
  • $\begingroup$ Ah thank you that explains it well for me! $\endgroup$
    – Danny King
    Mar 28, 2011 at 10:03

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

Not the answer you're looking for? Browse other questions tagged or ask your own question.