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Consider my poorly drawn triangle bounded by the three straight lines. All I've to do is to find the area of the traingle using integration.

But first of all I'll share what I know about finding the area using integration. Correct me if I'm wrong.

1)We consider small strips with length $f(x)$ and width $dx$ which is between the the curve and the corresponding x-axis or y-axis.

2)After that we integrate the equation of the curve, with limits ranging from the specific values of x-axis or y-axis and finally we get the area under the curve.

So I've split the triangle into two equal parts ABC and CBD.

But both parts are giving me problems, just consider ABC. To find the area, we integrate the equations of the straight lines 1 and 2 with limits 0 to 2 and the difference between the two will give us the area of ABC.

But there is a small part which I've encircled in the figure, the straight line there is below the x-axis, so doesn't that mean the strips does not exist there? If that is so, then won't it imply that the calculated value is less than the actual value? The same problem arises when finding the area of CBD. Please help me out of this problem.

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  • $\begingroup$ What if you integrated wrt y? $\endgroup$ – Andrew Li Feb 20 '18 at 5:01
  • $\begingroup$ @SlayerDiAngelo It might help to think that $\Delta ABD$ is the region between $AB$ and $AD$, minus the region between $AB$ and $BD$. $\endgroup$ – Toby Mak Feb 20 '18 at 5:01
  • $\begingroup$ The height of each strip is the difference of the two $y$-values, high $y$-value minus low $y$-value. Whether either or both $y$-values are negative is not relevant. High minus low, that's all. For example, if for a given strip, the high $y$-value is $3$, and the low $y$-value is $-5$, then the height of that strip is $3-(-5)=8$. $\endgroup$ – quasi Feb 20 '18 at 5:02
  • $\begingroup$ But the length of the strip is $f(x)$, so isn't it between the curve and the x-axis? $\endgroup$ – SlayerDiAngelo Feb 20 '18 at 5:08
  • $\begingroup$ No, the length of the strip is $f(x)-g(x)$ if $f(x) \ge g(x)$. Use a mental ruler to check. So the strip you drew (labeled $(1)$) is not correct, since it goes outside the triangle whose area you're trying to find. $\endgroup$ – quasi Feb 20 '18 at 5:26
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First, we should find equations for all of the lines. We can determine the slope of each line: $$m_{AB}=\frac{3-1}{2-0}=1$$ $$m_{BD} = \frac{-1-3}{4-2}= -2$$ $$m_{AD}=\frac{-1-1}{4-0}=-\frac{1}{2}$$ By using the slope-intercept form of a line and picking a point on each line, we see: $$\overline{AB}: y-1=1(x-0) \implies y=x+1$$ $$\overline{AD}:y-1=-\frac{1}{2}(x-0) \implies y=-\frac{1}{2}x+1 $$ $$\overline{BD}:y-3=-2(x-2) \implies y = -2x+7$$

Now we can begin our integration. If we wanted to integrate with respect to x, we can do so by noticing that for $0 \le x \le 2$, the region is bounded above by $\overline{AB}$ and for $2 \le x \le 4$, the region is bounded above by $\overline{BD}$. In both cases, the region is bounded below by $\overline{AD}$. This gives rise to our integral:

\begin{align} \\Area(\Delta ABD) & = \int_0^2 (x+1)-(-\frac{1}{2}x+1) \ dx +\int_2^4 (-2x+7) - (-\frac{1}{2}x + 1) \ dx \\ \\ & = \int_0^2 \frac{3}{2}x \ dx + \int_2^4 -\frac{3}{2}x+6 \ dx \\ \\ & = \left( \frac{3x^2}{4} \Biggr\rvert^2_0 \right) + \left(-\frac{3x^2}{4}+6x \Biggr\rvert^4_2 \right) \\ \\ & = (3 - 0) + (12 - 9) \\ \\ & = 6. \end{align}

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