2
$\begingroup$

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?

$\endgroup$
4
  • $\begingroup$ 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
    Commented 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
    Commented 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
    Commented 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
    Commented Mar 28, 2011 at 7:40

2 Answers 2

Reset to default
1
$\begingroup$

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).

$\endgroup$
1
$\begingroup$

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.

$\endgroup$
3
  • $\begingroup$ Interesting, thanks! In that case, how did you know this and also, was my method incorrect? (I got a different answer). $\endgroup$
    – Danny King
    Commented 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? $\endgroup$
    – Hans Lundmark
    Commented Mar 28, 2011 at 9:32
  • $\begingroup$ Ah thank you that explains it well for me! $\endgroup$
    – Danny King
    Commented Mar 28, 2011 at 10:03

You must log in to answer this question.

Not the answer you're looking for? Browse other questions tagged .