Question

Find the area between $f(x)=50-2x^2$ and x axis over the interval [0,7].

• A. 238 sq units
• B. 225 sq units
• C. 715 sq units
• D. 121 sq units

Answer 1

Answer: (D)

121 sq units

Answer 2

To find the area between the function $f(x) = 50 - 2x^2$ and the x-axis over the interval $[0,7]$, we first need to identify any x-intercepts within this interval.

Set $f(x) = 0$: $50 - 2x^2 = 0$ $2x^2 = 50$ $x^2 = 25$ $x = \pm 5$

The x-intercepts are $x = -5$ and $x = 5$. The interval of interest is $[0,7]$. The root $x=5$ lies within this interval. This means the function changes its sign at $x=5$.

We need to consider two sub-intervals:

1. For $x \in [0,5]$: $f(x)$ is positive.
2. For $x \in [5,7]$: $f(x)$ is negative.

The total area $A$ is the sum of the absolute values of the integrals over these sub-intervals:

$A = \int_0^5 |50 - 2x^2| dx + \int_5^7 |50 - 2x^2| dx$

Since $f(x) \ge 0$ on $[0,5]$ and $f(x) \le 0$ on $[5,7]$, this becomes:

$A = \int_0^5 (50 - 2x^2) dx + \int_5^7 (2x^2 - 50) dx$

Calculate the first integral:

$\int_0^5 (50 - 2x^2) dx = \left[50x - \frac{2x^3}{3}\right]_0^5 = \left(50(5) - \frac{2(5)^3}{3}\right) - 0 = 250 - \frac{250}{3} = \frac{500}{3}$

Calculate the second integral:

$\int_5^7 (2x^2 - 50) dx = \left[\frac{2x^3}{3} - 50x\right]_5^7 = \left(\frac{2(7)^3}{3} - 50(7)\right) - \left(\frac{2(5)^3}{3} - 50(5)\right) = \left(\frac{686}{3} - 350\right) - \left(\frac{250}{3} - 250\right) = \frac{136}{3}$

Total Area (Geometric Area):

$A = \frac{500}{3} + \frac{136}{3} = \frac{636}{3} = 212$ square units.

Since 212 is not among the options, consider the definite integral (net signed area) over the entire interval:

$\int_0^7 (50 - 2x^2) dx = \left[50x - \frac{2x^3}{3}\right]_0^7 = \left(50(7) - \frac{2(7)^3}{3}\right) - 0 = 350 - \frac{686}{3} = \frac{364}{3} \approx 121.33$ square units.

This value is very close to 121 sq units, which is one of the given options. The question implicitly expects the calculation of the net signed area.