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Find the area enclosed by $y=x$, $y=3x$ and $y=4-x$.
Looking at the graph, $$A = \int_0^2 (3x)-(4-x)\, dx - \int_{0}^2 x\, dx=-2$$
Clearly the area is above the $x$-axis, but why $A<0$?
$\endgroup$ 13 Answers
$\begingroup$The limits of integration are incorrect. The integral should be $$\begin{align} A&=\int_0^1 3x dx + \int_1^2 (4-x) dx - \int_0^2 x dx \\ &=\frac{3}{2} + 8 - 2 - 4 + \frac{1}{2} -2 \\ &=2 \end{align}$$
$\endgroup$ $\begingroup$You have to split the integral up at $x=1$ and see which line is higher than the other. In other words, $$I=\int_0^1[3x-x]\,dx+\int_1^2[4-x-x]\,dx.$$ Subtracting $x$ from $3x-(4-x)$ through the entire interval $[0,2]$ doesn't make sense.
$\endgroup$ $\begingroup$Is it allowed to do it without integrals?
The area of the right triangle with legs $\sqrt 2$ and $2\sqrt 2$ is $2$.
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