Find a constant so the area between two graphs has specified area

For a constant $a \in \mathbb{R}$ define

$f(x) = x - x^2, \qquad g(x) = ax.$

Find the values for $a$ such that the region above the graph of $g$ and below the graph of $f$ has area equal to $\frac{9}{2}$ .

We consider three cases: $a = 0, \ a < 0$ , and $a > 0$ .
Case 1: $a = 0$ . This is not possible since if $a = 0$ then $g(x) = ax = 0$ and so the area above the graph of $g$ and below the graph of $f$ is equal to

$\int_0^1 (x-x^2) \, dx = \left. \left( \frac{x^2}{2} - \frac{x^3}{3} \right) \right|_0^1 = \frac{1}{6} \neq \frac{9}{2}.$

Case 2: $a < 0$ . If $a < 0$ then $f(x) \geq g(x)$ on $[0,1-a]$ , so we have the area, $a(S)$ , of the region between the two graphs given by

$\begin{align*} a(S) = \int_0^{1-a} (x-x^2 - ax) \, dx &= (1-a) \int_0^{1-a} x \, dx - \int_0^{1-a} x^2 \, dx \\ &= (1-a) \left( \frac{(1-a)^2}{2} \right) - \frac{(1-a)^3}{3} \\ &= \frac{(1-a)^3}{6}. \end{align*}$

Setting this equal to $\frac{9}{2}$ and solving for $a$ we have

$\frac{(1-a)^3}{2} = \frac{9}{2} \quad \implies \quad (1-a)^3 = 27 \quad \implies \quad 1-a = 3 \quad \implies \quad a = -2.$

Case 3: $a > 0$ . If $a > 0$ then $f(x) \geq g(x)$ on $[1-a,0]$ so

$\begin{align*} a(S) = \int_{1-a}^0 (x-x^2-ax)\, dx &= (1-a)\int_{1-a}^0 x \, dx - \int_{1-a}^0 x^2 \, dx \\ &= (1-a) \left( \frac{-(1-a)^2}{2} - \left( \frac{(1-a)^3}{2} \right) \\ &= - \frac{(1-a)^3}{6}. \end{align*}$

Setting this equal to $\frac{9}{2}$ and solving for $a$ we obtain

$-\frac{(1-a)^3}{6} = \frac{9}{2} \quad \implies \quad (1-a)^3 = -27 \quad \implies \quad a = 4.$

Thus, the possible values of $a$ to make the area of the region above the graph of $g$ and below the graph of $f$ are $-2, 4$ .

3 comments

July 8, 2023 at 5:57 am

Bruno Lossada says:

Hi! First of all, thank you so much for your help with these solutions! I have a quick question about this exercise. The statement says ‘the region above the graph of g and below the graph of f.’ Then, shouldn’t we assume the case f > g?

April 17, 2016 at 4:23 pm

Edwin says:

Hi! Quick question, how did you know that the interval was $[ 0,1-a]$ when $a < 0$ ?

- April 17, 2016 at 4:59 pm
  
  RoRi says:
  
  Hi! Since we want to look at the area below $f(x)$ and above $g(x)$ we want the interval to correspond to where $g(x) \leq f(x)$ . So, we look for
  
  $\begin{align*} && f(x) - g(x) &\geq 0 \\ \implies && x - x^2 - ax &\geq 0 \\ \implies && (1-a)x - x^2 & \geq 0 \\ \implies && (1-a)x &\geq x^2. \end{align*}$
  
  We have equality if $x = 0$ . If $x \neq 0$ then this gives us $x \leq (1-a)$ . Since we have $a < 0$ in this case by assumption, we know $1-a > 0$ so the interval is $[0, 1-a]$ . So we have $f(x) \geq g(x)$ when $0 \leq x \leq 1-a$ .

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Find a constant so the area between two graphs has specified area

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