# Area between the curve $y$, the $x$-axis and the lines

Find the total area between the curve $y=x^2-4x+3$, the $x$-axis and the lines $x=0$ and $x=3$.

I have drawn the graph and concluded that:

$$\int_0^1 0 - (x^2-4x+3)\, dx + \int_1^3 0 - (x^2-4x+3)\, dx = -\frac{-26}{3}$$ Which differs from the answer key: $4$. What is the mistake here? Is it feasiable to solve this problem without relying on the graph?

• the first integral should be positive, because the upper graphic is $x^2-4x+3$ – haqnatural Aug 31 '18 at 4:04

The required area is the green area:

$\hspace{3cm}$

The proper set up of integrals: $$\int_0^1 (x^2-4x+3)-0\, dx + \int_1^3 0 - (x^2-4x+3)\, dx = \frac83.$$

$\int_0^1 (x^2-4x+3)\mathbb dx-\int_1^3(x^2-4x+3)\mathbb dx=[(\frac{x^3}3-2x^2+3x)]_0^1-[\frac{x^3}3-2x^2+3x]_1^3=\frac43-[0-\frac43]=\frac83$

Hint:

Area under curve of $y=f(x)$ in the domain of $a < x < b$ is given by \begin{align} \mathrm{Area} \; = \; \int_{a}^{b} \Big|\, f(x) \, \Big| \; dx \end{align} Take absolute value of your integrand.

For your case, $f(x) = x^{2} - 4x + 3$. Factorization gives $f(x) = (x-3)(x-1)$. Now you can easily see when $f(x) > 0$ or $<0$ without using a graph.