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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?

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    I think "lies on the +ve x-axis" in my comment above should read "lies in the 1st quadrant"2011-03-28

2 Answers 2

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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).

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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.

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    Ah thank you that explains it well for me!2011-03-28