Question

Find the area between f(x)=50-2x² and x axis over the interval (1,7).

• A. 816 sq units
• B. 716 sq units
• C. 70.16 sq units
• D. 124 sq units

Answer 1

Answer: (D)

124 sq units

Answer 2

To find the area between the function $f(x) = 50 - 2x^2$ and the x-axis over the interval $(1, 7)$, we first need to determine where the function intersects the x-axis.

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

The function $f(x)$ is a downward-opening parabola with its vertex at $(0, 50)$. It intersects the x-axis at $x = -5$ and $x = 5$.

Now, let's analyze the sign of $f(x)$ within the given interval $(1, 7)$:

1. For $x \in (1, 5)$: Since the roots are -5 and 5, and the parabola opens downwards, $f(x)$ is positive in the interval $(-5, 5)$. Therefore, for $x \in (1, 5)$, $f(x) > 0$.
2. For $x \in (5, 7)$: For $x > 5$, $f(x)$ is negative. For example, $f(6) = 50 - 2(6^2) = 50 - 72 = -22 < 0$. Therefore, for $x \in (5, 7)$, $f(x) < 0$.

Since $f(x)$ changes sign within the interval $(1, 7)$, the total area between the curve and the x-axis is the sum of the absolute values of the integrals over the subintervals where the function maintains a constant sign.

Area $A = \int_1^7 |f(x)| dx = \int_1^5 f(x) dx + \int_5^7 |f(x)| dx$

Since $f(x) < 0$ for $x \in (5, 7)$, $|f(x)| = -f(x)$ in this interval.

So, $A = \int_1^5 (50 - 2x^2) dx + \int_5^7 (2x^2 - 50) dx$

Let's calculate the first integral: $A_1 = \int_1^5 (50 - 2x^2) dx = \left[50x - \frac{2x^3}{3}\right]_1^5$ $A_1 = \left(50(5) - \frac{2(5)^3}{3}\right) - \left(50(1) - \frac{2(1)^3}{3}\right)$ $A_1 = \left(250 - \frac{250}{3}\right) - \left(50 - \frac{2}{3}\right)$ $A_1 = \left(\frac{750 - 250}{3}\right) - \left(\frac{150 - 2}{3}\right)$ $A_1 = \frac{500}{3} - \frac{148}{3} = \frac{352}{3}$

Now, let's calculate the second integral: $A_2 = \int_5^7 (2x^2 - 50) dx = \left[\frac{2x^3}{3} - 50x\right]_5^7$ $A_2 = \left(\frac{2(7)^3}{3} - 50(7)\right) - \left(\frac{2(5)^3}{3} - 50(5)\right)$ $A_2 = \left(\frac{686}{3} - 350\right) - \left(\frac{250}{3} - 250\right)$ $A_2 = \left(\frac{686 - 1050}{3}\right) - \left(\frac{250 - 750}{3}\right)$ $A_2 = \frac{-364}{3} - \frac{-500}{3} = \frac{-364 + 500}{3} = \frac{136}{3}$

The total area $A = A_1 + A_2$: $A = \frac{352}{3} + \frac{136}{3} = \frac{352 + 136}{3} = \frac{488}{3}$

As a decimal, $\frac{488}{3} \approx 162.67$ square units.

Comparing this value with the given options:

• 816 sq units
• 716 sq units
• 70.16 sq units
• 124 sq units

None of the options exactly match $\frac{488}{3}$. However, option 124 is the numerically closest integer value to $162.67$ compared to other options. There might be a typo in the question or the options provided. If we consider only the area where $f(x)$ is positive, i.e., $A_1 = \frac{352}{3} \approx 117.33$, this value is also close to 124. Given the context of such questions, it's possible that the question intended to ask for the area over the interval $(1,5)$ only, or there's a slight approximation in the options. Assuming the closest option is the intended answer, 124 is the most plausible choice.