Area enclosed between two curves equal to 5.

Question

Prove that there exists a ∈ R so that the area bound by the curves $y = x^2$ and y=ax+1 is equal to 5.

I was trying to solve this question but I can't really seem to understand how to approach it since there is no information on how the curves will intersect.

Answer 1

Although @RossMillikan gave you the answer, i'll add the specific calculation for the value of $a$ which solves your question.

First thing first you have to find the points of intersection of the parabola with the line. They can easily be found solving the equation

$$ x^2 = ax+1 \implies x^2-ax-1 = 0 \implies x_{1,2} = \frac{+a\pm\sqrt{a^2+4}}{2} $$

Now if you look at a graph of a line intersecting a parabola it's easy to see that the area bound by the two curves is just the area under the line minus the area under the parabola between the two points of intersection. Using simple integration

$$\int\limits_{\frac{+a-\sqrt{a^2+4}}{2}}^{\frac{+a+\sqrt{a^2+4}}{2}} [(ax+1)-x^2]dx = \left.a\frac{x^2}{2} + x - \frac{x^3}{3}\right|_{\frac{+a-\sqrt{a^2+4}}{2}}^{\frac{+a+\sqrt{a^2+4}}{2}} = \frac{1}{6}(a^2+4)^{3/2}$$

Now you can just put this equal to $5$ and find the asked value of $a$.

At the end you'll find two values for $a$ belonging to the two symmetric lines with respect to the $y$-axis, namely

$$a= \pm\sqrt{30^{2/3}-4} $$

Answer 2

Draw a picture for $a=0, 1, 5$. It should have one parabola from $y=x^2$ and three lines on it. You will see how the curves intersect. If you compute the area for $a=0$ you should find it is less than $5$. Intuitively, if $a$ is very large, the area will be, too. By the intermediate value theorem there is some $a$ that makes the area $5$. You are not asked to find the required $a$, though it would not be too hard. You just have to justify the statement that begins "Intuitively".

Answer 3

There are two ways of doing this:

1. Find the particular value of a when the area is 5

2. Find a value of area where the area is less than 5, find another value where it is greater than 5, and then show on the interval in between that the area changes continuously with a

Answer 4

HINT

Broad method indicated, please check it.

Real roots are $ a/2 \pm \sqrt{(a/2)^2-1} = (q,p)$ say

$$\int_p^q(x^2-p x-q x +p q) dx$$

$$ = (q-p)(p^2+5q^2)/6$$

Plug-in and simplify