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.
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}