Zero net area Consider the function ƒ(𝓍) = 𝓍² ― 4𝓍 .                                                                                                                                       
                                                                                                                                                                                     c) In general, for the function ƒ(𝓍) = 𝓍² ― a𝓍, where a > 0, for what value of b > 0 (as a function of a) is ∫₀ᵇ ƒ(𝓍) d𝓍 = 0 ? 

Verified step by step guidance

1. Start by writing down the integral you need to solve: \(\int_0^b (x^2 - a x) \, dx = 0\), where \(a > 0\) and \(b > 0\).
2. Compute the indefinite integral of the function \(f(x) = x^2 - a x\). The antiderivative is \(\int (x^2 - a x) \, dx = \frac{x^3}{3} - \frac{a x^2}{2} + C\).
3. Evaluate the definite integral from 0 to \(b\) using the antiderivative: \(\left[ \frac{x^3}{3} - \frac{a x^2}{2} \right]_0^b = \frac{b^3}{3} - \frac{a b^2}{2} - \left(0\right)\).
4. Set the definite integral equal to zero to find \(b\): \(\frac{b^3}{3} - \frac{a b^2}{2} = 0\).
5. Factor the equation to solve for \(b\): \(b^2 \left( \frac{b}{3} - \frac{a}{2} \right) = 0\). Since \(b > 0\), solve \(\frac{b}{3} - \frac{a}{2} = 0\) for \(b\) to express \(b\) as a function of \(a\).

Key Concepts

Definite Integral and Net Area

The definite integral of a function over an interval represents the net area between the function's graph and the x-axis. Positive areas above the x-axis add to the integral, while areas below subtract. When the integral equals zero, the positive and negative areas cancel out, resulting in zero net area.

Finding the Integral of a Quadratic Function

To evaluate the integral of a quadratic function like ƒ(x) = x² - a x, you apply the power rule for integration term-by-term. This involves increasing the exponent by one and dividing by the new exponent, then applying limits to find the definite integral value as a function of the upper limit b.

Solving for the Upper Limit to Achieve Zero Net Area

Setting the definite integral equal to zero and solving for the upper limit b involves forming an equation from the integral expression and isolating b. This process finds the point where the accumulated positive and negative areas balance, which depends on the parameter a in the function.