Problem 68 - Calculus for Biology and Medicine

Question

$$ \text { Find } a>0 \text { such that } \int_{0}^{a}(1-x) d x=0 $$

Step-by-Step Solution

Answer

The value of \( a \) is 2.

Step 1: Set up the integral

The problem asks us to find \( a > 0 \) such that the integral of \( 1-x \) from 0 to \( a \) is zero. We start by writing the definite integral as \(\int_{0}^{a} (1-x) \, dx = 0. \)

Step 2: Compute the integral

To solve the integral, we find the antiderivative of \( 1-x \), which is \( x - \frac{x^2}{2} \). So:\[\int (1-x) \, dx = x - \frac{x^2}{2} + C.\]

Step 3: Evaluate the integral from 0 to a

Now, evaluate the antiderivative from \( 0 \) to \( a \):\[\left. \left(x - \frac{x^2}{2} \right) \right|_0^a = \left(a - \frac{a^2}{2} \right) - \left(0 - \frac{0^2}{2} \right).\]Thus, the result is \( a - \frac{a^2}{2} \).

Step 4: Set the result equal to zero

According to the problem, this expression should be equal to zero:\[a - \frac{a^2}{2} = 0.\]

Step 5: Solve the equation for a

Rearrange the equation:\[a - \frac{a^2}{2} = 0 \rightarrow a = \frac{a^2}{2}.\]Multiply all terms by 2 to clear the fraction:\[2a = a^2.\]Rearrange: \( a^2 - 2a = 0 \).Factor the equation: \( a(a - 2) = 0 \).This gives solutions: \( a = 0 \) and \( a = 2 \).

Step 6: Select the positive value of a

Since the problem specifies \( a > 0 \), we select \( a = 2 \) as the solution.

Key Concepts

Antiderivative

An antiderivative is the reverse process of differentiation. It is a function whose derivative is the original function. For example, if we take the function \(1 - x\), its derivative is \(-1\) with respect to \(x\). Thus, an antiderivative of \(1-x\) can be determined by finding a function whose derivative gives us \(1-x\).
For the equation \(1-x\), the antiderivative is \(x - \frac{x^2}{2} + C\), where \(C\) represents the constant of integration. This means if you differentiate \(x - \frac{x^2}{2}\), you will obtain \(1-x\).
Calculating the antiderivative is a crucial step in solving integrals. By determining it, we convert a once complex integral into an expression we can easily evaluate at the boundaries of the specified interval.

Evaluate Integral

Evaluating an integral involves substituting the upper and lower bounds into the antiderivative. This process helps us find the net area under a curve between two points.
Given the problem, first, compute the integral \(\int_{0}^{a}(1-x) \, dx\) by using its antiderivative, which we've determined as \(x - \frac{x^2}{2}\).
The Fundamental Theorem of Calculus tells us to evaluate this antiderivative from the lower limit (0) to the upper limit (\(a\)): \ \[\left.\left(x - \frac{x^2}{2} \right)\right|_0^a = \left(a - \frac{a^2}{2} \right) - \left(0 - \frac{0^2}{2}\right)\]
Replacing \(x\) with \(a\) gives \(a - \frac{a^2}{2}\) while replacing \(x\) with 0 simplifies to 0. Ultimately, the expression becomes \(a - \frac{a^2}{2}\), representing the exact net area we want to equate to zero to find \(a\).

Solve Equation

To solve for \(a\) that satisfies the condition of the integral, set the evaluated integral equal to zero:
\[a - \frac{a^2}{2} = 0\]
Rearranging gives \(a = \frac{a^2}{2}\). Multiplying every term by 2 simplifies it to \(2a = a^2\).
Now, further rearranging leads to the factorable quadratic equation:

• \(a^2 - 2a = 0\)

By factoring, we find:
\[a(a-2) = 0\]
This equation implies two possible solutions for \(a\):

• \(a = 0\)
• \(a = 2\)

However, given \(a > 0\) as a condition in the problem, we conclude \(a = 2\) as the valid solution. This process not only finds the value of \(a\) but also broadens the understanding of solving quadratic equations resulting from definite integrals.