1
$\begingroup$

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?

$\endgroup$
  • $\begingroup$ the first integral should be positive, because the upper graphic is $x^2-4x+3$ $\endgroup$ – haqnatural Aug 31 '18 at 4:04
1
$\begingroup$

The required area is the green area:

$\hspace{3cm}$enter image description here

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

$\endgroup$
1
$\begingroup$

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

$\endgroup$
0
$\begingroup$

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.

$\endgroup$

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

Not the answer you're looking for? Browse other questions tagged or ask your own question.