Problem 83 - Calculus for Biology and Medicine

Question

Find \(a>1\) such that $$ \int_{1}^{a}(x-2)^{3} d x=0 $$

Step-by-Step Solution

Answer

The value of \( a \) is 3.

Step 1: Identify the Integral to Solve

We need to find \( a > 1 \) such that the definite integral \( \int_{1}^{a}(x-2)^{3} \, dx = 0 \). This integral represents the area under the curve \( (x-2)^3 \) from 1 to \( a \).

Step 2: Compute the Indefinite Integral

First, compute the indefinite integral of \( (x-2)^3 \). Using the power rule for integration, which states that \( \int x^n \, dx = \frac{x^{n+1}}{n+1} + C \), we integrate with respect to \( x \):\[ \int (x-2)^3 \, dx = \frac{(x-2)^4}{4} + C \].

Step 3: Evaluate the Definite Integral

Using the result from the indefinite integral, evaluate the definite integral from 1 to \( a \):\[ \int_{1}^{a}(x-2)^3 \, dx = \left[ \frac{(x-2)^4}{4} \right]_{1}^{a} \].Substitute the limits of integration into this expression:\[ \frac{(a-2)^4}{4} - \frac{(1-2)^4}{4} = 0 \].

Step 4: Solve the Resulting Equation

Based on the previous evaluation, we have the equation:\[ \frac{(a-2)^4}{4} - \frac{1}{4} = 0 \].Multiplying through by 4 to eliminate the fraction yields:\[ (a-2)^4 - 1 = 0 \].Rewriting it gives:\[ (a-2)^4 = 1 \].

Step 5: Find the Value of \( a \)

To solve \( (a-2)^4 = 1 \), take the fourth root of both sides:\[ a-2 = 1 \].Adding 2 to both sides, we find:\[ a = 3 \].Therefore, the value of \( a \) satisfying \( a > 1 \) is 3.

Key Concepts

Definite Integral

In calculus, a definite integral is used to compute the signed area under a curve on a given interval, usually denoted as \[ \int_{a}^{b} f(x) \, dx \].The bounds of integration, \(a\) and \(b\), determine the segment of the curve where the area is calculated. For instance, the definite integral\[ \int_{1}^{a}(x-2)^{3} \, dx = 0 \]in the original exercise computes the area under the curve \((x-2)^3\) from \(x = 1\) to \(x = a\).

• **Evaluating Areas**: The definite integral results in a numerical value representing this area.
• **Positive and Negative Areas**: Depending on the curve's position relative to the x-axis, the area can be positive (above the x-axis), negative (below the x-axis), or even zero.

In the example given, setting the integral equal to zero helps find an upper limit of integration \(a\) where the accumulated area from 1 to \(a\) equals zero. The solution showed that \(a = 3\) is such a point, resulting in both positive and negative areas canceling each other out.

Indefinite Integral

The indefinite integral, represented as \[ \int f(x) \, dx \],provides a family of functions whose derivative is the integrand. A general solution involves an arbitrary constant \(C\), making it more general. In the example, the indefinite integral is of the form \[ \int (x-2)^3 \, dx \].

• **Power Rule**: This integral is solved using the power rule, stating \( \int x^n \, dx = \frac{x^{n+1}}{n+1} + C \).
• **Result**: Applying this rule to \((x-2)^3\) yields the antiderivative: \( \frac{(x-2)^4}{4} + C \).

The constant \(C\) is fundamental in indefinite integrals, accounting for all possible vertical shifts of the antiderivative graph. However, this constant does not affect definite integrals as the constant cancels out during computation using upper and lower limits.

Integration Techniques

Integration techniques are critical tools in calculus that help solve integrals efficiently. Various techniques can be employed depending on the function being integrated:

• **Power Rule**: As seen in the example, it's suitable for polynomials, where you increase the power by one and divide by the new power.
• **Substitution**: This technique is useful when you can transform the integrand to a simpler form by re-casting the variable.
• **Integration by Parts**: Helpful for products of functions, following the formula \( \int u \, dv = uv - \int v \, du \).
• **Partial Fractions**: Beneficial when dealing with rational functions, breaking down complex fractions into simpler terms.

These techniques are the foundation of solving real-world calculus problems effectively and allow for tackling a broad spectrum of integral equations. Mastery of these methods is crucial for advancing in higher mathematics and practical applications. In the current exercise, the power rule was used effectively to solve for the indefinite integral of \((x-2)^3\).