Areas and integrals

Calculus Level 3

A curve has equation $y=f(x)$ , where $f(x)=x(x-p)(x-q)(r-x)$ and $0<p<q<r$ .

If $\int \limits_0^rf(x)\, dx=0$ , $\int \limits_0^qf(x)\, dx=-2$ and $\int \limits_p^r f(x)\, dx=-3$ , what is the area enclosed between the curve and the $x$ -axis for $0\leqslant x\leqslant r$ ?

Clarification

$0\leqslant x\leqslant r\iff x\in [0,r]$

6 10 4 5 0 1

1 solution

Matthew Christopher
Aug 12, 2020

The curve looks something like this:

Call the area of the regions enclosed by the curve and the x-axis $A, B$ and $C$ , from left to right.

An area above the x-axis contributes positively to an integral, and an area below it contributes negatively.

Hence $\begin{cases}A-B+C=0\\A-B=-2\\-B+C=-3\end{cases}$

Solving these simultaneously, $A=3$ , $B=5$ and $C=2$ .

The desired area $=A+B+C=3+5+2=\color{#20A900}{\boxed{10}}$ .

From the graph, it looks like

The area of A-B Is the same as -B+C

Change the graph

Vijay Simha - 10 months ago

Thank you for the suggestion. I have changed the graph, as suggested.

Matthew Christopher - 10 months ago

At first I didn't understand your question because I didn't know that notation to indicate intervals (if it is what you mean). I thought it was a greater or equal than symbol. To make it more clear I think you should change the notation to $[0, r]$ as I think it is a more standard notation.

Saúl Huerta - 9 months, 4 weeks ago

Yes, $0\leqslant x\leqslant r$ is equivalent to $x\in [0,r]$ . It is, in fact, a greater than or equal to symbol. Sorry for the confusion.

Matthew Christopher - 9 months, 4 weeks ago

Why did you define $f(x)$ as $f(x)=x(x-p)(x-q)(r-x)$ tho? I think it's redundant.

We have:

$\int_a^b f(x) \, dx = \int_a^c f(x) \, dx + \int_c^b f(x) \, dx$

For any choice of f

Halim Amran - 6 months ago

Areas and integrals

1 solution

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