Find the area between the curve and the x-axis

[b4rn5ey]

Morning all,

Got some feedback on some recent work I submitted, and I've only gone wrong on one calculation (Woo!) - however I have no idea where for this one question.

The Question is as follows:

Find the area between the curve y=x² - x - 2 and x-axis in the range y=-3 to x=5.

Here is how I went about it originally:

Imgur: The most awesome images on the Internet

The tutor has advised: Total area requires correction. You should first identify points of intersection with the x-axis then evaluate the area of several parts changing the negative area to positive.

Frankly, he's lost me..haha

Any help would be greatly appreciated

 

[Opalg]

Morning all,

Got some feedback on some recent work I submitted, and I've only gone wrong on one calculation (Woo!) - however I have no idea where for this one question.

The Question is as follows:

Find the area between the curve y=x² - x - 2 and x-axis in the range y=-3 to x=5.

Here is how I went about it originally:

Imgur: The most awesome images on the Internet

The tutor has advised: Total area requires correction. You should first identify points of intersection with the x-axis then evaluate the area of several parts changing the negative area to positive.

Frankly, he's lost me..haha

Any help would be greatly appreciated

As your sketch shows, the curve lies below the $x$-axis in the interval $x=-1$ to $x=2$. When you integrate, areas below the axis count as negative. But the question (apparently) wants you to count such areas as positive. So instead of calculating a single integral $$\int_{-3}^5(x^2 - x - 2)\,dx$$, you need to find three separate integrals $$\int_{-3}^{-1}(x^2 - x - 2)\,dx$$, $$\int_{-1}^2(x^2 - x - 2)\,dx$$ and $$\int_{2}^5(x^2 - x - 2)\,dx$$. The first and third of these will be positive. But the second one will give a negative result, and you will need to change the sign to make it positive, before adding it to the other two to get the final result.

 

[skeeter]

note ...

$$|x^2-x-2|=|(x+1)(x-2)|=\left\{\begin{matrix}
x^2-x-2 &; \, x \le-1 \\
-(x^2-x-2) &; \, -1<x<2 \\
x^2-x-2& ; \, x \ge 2
\end{matrix}\right.$$

$$A=\int_{-3}^5 |x^2-x-2| \,dx = \int_{-3}^{-1} (x^2-x-2) \,dx - \int_{-1}^{2} (x^2-x-2) \,dx + \int_{2}^{5} (x^2-x-2) \,dx$$

... as stated by Opalg