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I encountered this problem and did not know how to solve it:

enter image description here

I graphed it out - making a point for each vertice. I then decided there must be a 3 because when $x=0$ then $y=3$. But, according to that logic, it should be answer c - because when $y=0$ then $x=3$. But, that is not the answer.

Where did I go wrong, and how should I go about solving this problem when I am not given an answer key?

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    $\begingroup$ I'm not sure what the problem is. Isn't the answer actually (c)? (Normally, we would write that $\int_{x=0}^3 (3+x)\,dx$ to make it clearer what the integrand is, but I don't think that's what the issue is here.) $\endgroup$ – Brian Tung Jul 22 at 3:39
  • $\begingroup$ In the picture, didn't you click option d)? $\endgroup$ – Toby Mak Jul 22 at 3:42
  • $\begingroup$ Yes, I did. But, why is the answer not answer d? $\endgroup$ – burt Jul 22 at 3:45
  • $\begingroup$ Well, is $3-x$ equal to $6$ when $x = 3$? If not, then (d) is not correct. (Remember that $(3, 6)$ represents $x = 3, y = 6$.) $\endgroup$ – Brian Tung Jul 22 at 3:46
  • $\begingroup$ But, when $y=0$ then $x$ is not equal to $-3$. Or, am I just getting majorly confused? $\endgroup$ – burt Jul 22 at 3:48
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When you look at the picture of the trapezoid in question, to find its area you are actually integrating a line segment from $(0,3)$ to $(3,6)$ in $dx$.

The slope of this is line is $(6-3)/(3-0) = 1$ and plugging in the first point yields the equation $y=x+3$. Thus we will integrate $(x+3)$.

As for the limits, you need to go $(0,3) \to (3,6)$ in $x$, so $x=0$ to $x=3$...

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  • $\begingroup$ So I should find the two points that really include the area of the entire thing, find the equation of that line, and then take the integral of that? $\endgroup$ – burt Jul 22 at 3:54
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    $\begingroup$ @burt that's the idea, when you integrate an area under some function... $\endgroup$ – gt6989b Jul 22 at 14:53
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Well, the correct answer is c. The slope of the line that represents the upper part of the trapezoid should be $m=\frac{3}{3}=1$ (the run is $3$ and the rise is also $3$ because $6-3=3$) and the $y$-intercept of that line, $b$, is $3$. Plugging all that information into the slop-intercept form of a line $y=mx+b$ gives us:

$$ y=x+3. $$

And to get the area under that curve, you would integrate it from $0$ to $3$:

$$ \int_{0}^{3}(x+3)\,dx. $$

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The equation of the segment joining $ (0,3)$ to $(3,6)$ is $$y=3+x$$

You are finding the area under $$y=3+x$$ and over the $x$-axis from $x=0$ to $x=3$

The correct choice is $(c)$

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  • $\begingroup$ So I first find the equation of the line that the area under the line is the area of the trapezoid? $\endgroup$ – burt Jul 22 at 4:04
  • $\begingroup$ Yes, of course because that is the one which you are going to integrate. $\endgroup$ – Mohammad Riazi-Kermani Jul 22 at 4:17
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The area is $\int_0^3 (x+3)dx = \frac{27}{2} = 13.5$

This is the area between the curve $y = x+3$ and the $x$-axis, between $x = 0$ and $x = 3$. You are trying to calculate this area, right?

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