Finding the area bounded by $y = 2 {x} - {x}^2 $ and straight line $ y = - {x}$

Question

$$ y =\ 2\ {x} - {x}^2 $$ $$ y =\ -{x} $$

According to me , the area

$$ \int_{0}^{2}{2x\ -\ { x} ^2}\, dx \ + \int_{2}^{3}{\ {x} ^2\ -\ 2{x} }\, dx \\ $$

Which gives the area $ \frac{8}{3}$

But the answer is $ \frac{9}{2}$

Answer 1

The first graph is always on top for $x \in (0,3)$, it should be

$$\int_0^3 (2x-x^2) - (-x) \, dx$$

You have computed the following region instead.

Answer 2

By a sketch we can see that the correct set up is

$$\int_{0}^{3}{[(2x\ -\ { x} ^2)-(-x)]}\, dx=\int_{0}^{3}{(3x\ -\ { x} ^2)}\, dx$$

Answer 3

Equate $-x = 2x-x^2$

Points of intersection is $x=0$ and $x = 3$

Now area bounded by the curves is

$$\int_{0}^{3}(2x-x^2-(-x))dx = \int_{0}^{3} (3x-x^2)dx = \frac{9}{2}$$

Answer 4

To find the points of intersection between the curve $y = 2x - x^2$ and $y = -x$, equate the two expressions. \begin{align*} 2x - x^2 & = -x\\ 3x - x^2 & = 0\\ x(3 - x) & = 0 \end{align*} which has solutions $x = 0$ and $x = 3$. These are the limits of integration.

Since \begin{align*} y & = 2x - x^2\\ & = -x^2 + 2x\\ & = -(x^2 - 2x)\\ & = -(x^2 - 2x + 1) + 1\\ & = -(x - 1)^2 + 1 \end{align*} the graph of $y = 2x - x^2$ is a parabola with vertex $(1, 1)$ that opens downwards. Hence, it is above the line $y = -x$ in the interval $(0, 3)$, as shown in the figure below.

We want to find the area of the shaded region. Since it lies below the parabola and above the line, its area is $$\int_{0}^{3} [2x - x^2 - (-x)]~\textrm{d}x = \int_{0}^{3} (3x - x^2)~\textrm{d}x = \frac{9}{2}$$

Answer 5

You should go around the path in clockwise fashion: $$ \int_0^3(2x-x^2)\,dx+\int_3^0(-x)\,dx= \int_0^3(2x-x^2+x)\,dx=\Bigl[\frac{3}{2}x^2-\frac{1}{3}x^3\Bigr]_0^3=\frac{27}{2}-9=\frac{9}{2} $$