3
$\begingroup$

Find the area enclosed by $y=x$, $y=3x$ and $y=4-x$.

Looking at the graph, $$A = \int_0^2 (3x)-(4-x)\, dx - \int_{0}^2 x\, dx=-2$$

enter image description here

Clearly the area is above the $x$-axis, but why $A<0$?

$\endgroup$
1
  • 1
    $\begingroup$ You need to break it up into two regions: $$A=\int_0^1 (3x-x)\,dx+\int_1^2(4-x-x)\,dx. $$ $\endgroup$ Aug 31, 2018 at 15:48

3 Answers 3

1
$\begingroup$

The limits of integration are incorrect. The integral should be $$\begin{align} A&=\int_0^1 3x dx + \int_1^2 (4-x) dx - \int_0^2 x dx \\ &=\frac{3}{2} + 8 - 2 - 4 + \frac{1}{2} -2 \\ &=2 \end{align}$$

$\endgroup$
1
$\begingroup$

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.

$\endgroup$
0
$\begingroup$

Is it allowed to do it without integrals?

The area of the right triangle with legs $\sqrt 2$ and $2\sqrt 2$ is $2$.

$\endgroup$

You must log in to answer this question.

Not the answer you're looking for? Browse other questions tagged .