Problem 50 - Calculus for Biology and Medicine

Question

Given that \(\int_{0}^{u} x^{2} d x=\frac{1}{3} a^{3}\) evaluate the following: (a) \(\int_{0}^{1} \frac{1}{2} x^{2} d x\) (b) \(\int_{0}^{-1} 3 x^{2} d x\) (c) \(\int_{-1}^{2} \frac{1}{3} x^{2} d x\) (d) \(\int_{1}^{1} 3 x^{2} d x\) (e) \(\int_{-2}^{3}(x+1)^{2} d x\) (f) \(\int_{2}^{4}(x-2)^{2} d x\)

Step-by-Step Solution

Answer

(a) \( \frac{1}{6} \), (b) \( 1 \), (c) \( 1 \), (d) \( 0 \), (e) \( \frac{65}{3} \), (f) \( 24 \).

Step 1: Understand the Given Information

The given integral \( \int_{0}^{u} x^{2} d x = \frac{1}{3} a^{3} \) means the integral of \( x^2 \) from 0 to \( u \) equals \( \frac{1}{3} \) times \( a^3 \). We know \( \int x^2 \, dx = \frac{x^3}{3} + C \). The formula implies that \( a \) must be \( u \), so by integrating with known bounds, comparison can be used for substituting values.

Step 2: Compute the Integral for (a)

The given integral is \( \int_{0}^{1} \frac{1}{2} x^{2} d x \). To solve, compute \( \frac{1}{2} \times \int_{0}^{1} x^2 \, dx \). The antiderivative of \( x^2 \) is \( \frac{x^3}{3} \), so \[ \int_{0}^{1} x^2 \, dx = \left[ \frac{x^3}{3} \right]_0^1 = \frac{1^3}{3} - \frac{0^3}{3} = \frac{1}{3}. \] Thus, \[ \int_{0}^{1} \frac{1}{2} x^2 \, dx = \frac{1}{2} \times \frac{1}{3} = \frac{1}{6}. \]

Step 3: Evaluate Integral (b)

The integral is \( \int_{0}^{-1} 3 x^{2} d x \), evaluated from 0 to -1. Switch limits of integration and change the sign: \[ \int_{-1}^{0} 3 x^2 \, dx = 3 \times \left[ \frac{x^3}{3} \right]_{-1}^0 = \left[ x^3 \right]_{-1}^0. \] Compute: \( 0^3 - (-1)^3 = 0 + 1 = 1 \). Thus, the solution is \( 1 \).

Step 4: Solve Integral (c)

This integral is \( \int_{-1}^{2} \frac{1}{3} x^{2} d x \). \[ \frac{1}{3} \times \int_{-1}^{2} x^2 \, dx = \frac{1}{3} \times \left[ \frac{x^3}{3} \right]_{-1}^2 = \frac{1}{3} \times \left( \frac{2^3}{3} - \frac{(-1)^3}{3} \right) \]. Substitute and simplify: \[ \frac{1}{3} \times \left( \frac{8}{3} + \frac{1}{3} \right) = \frac{1}{3} \times \frac{9}{3} = \frac{1}{3} \times 3 = 1. \]

Step 5: Solve Integral (d)

This integral \( \int_{1}^{1} 3 x^2 d x \) is from 1 to 1, where the lower and upper limits are equal. When the limits are the same, the integral is zero: \( 0 \).

Step 6: Solve Integral (e)

For \( \int_{-2}^{3}(x+1)^2 d x \), expand \( (x+1)^2 = x^2 + 2x + 1 \). Thus, \[ \int (x^2 + 2x + 1) \, dx = \left[ \frac{x^3}{3} + x^2 + x \right]_{-2}^{3}. \] Evaluate: \[ \left( \frac{3^3}{3} + 3^2 + 3 \right) - \left( \frac{(-2)^3}{3} + (-2)^2 + (-2) \right) = (9 + 9 + 3) - \left( \frac{-8}{3} + 4 - 2 \right). \] Simplify: \( 21 - \left( -\frac{8}{3} + 2 \right) = 21 + \frac{8}{3} - 2 = 19 + \frac{8}{3} = \frac{57}{3} + \frac{8}{3} = \frac{65}{3} \).

Step 7: Solve Integral (f)

For \( \int_{2}^{4}(x-2)^2 d x \), expand \( (x-2)^2 = x^2 - 4x + 4 \). Thus, \[ \int (x^2 - 4x + 4) \, dx = \left[ \frac{x^3}{3} - 2x^2 + 4x \right]_{2}^{4}. \] Evaluate at the bounds: \[ \left( \frac{4^3}{3} - 2 \times 4^2 + 4 \times 4 \right) - \left( \frac{2^3}{3} - 2 \times 2^2 + 4 \times 2 \right). \] Simplify: \( \left( \frac{64}{3} - 32 + 16 \right) - \left( \frac{8}{3} - 8 + 8 \right) = \left( \frac{64}{3} - 16 \right) - \frac{8}{3} = \frac{64}{3} - 16 - \frac{8}{3} = \frac{56}{3} - \frac{24}{1} = \frac{56}{3} - \frac{72}{3} = 24 \).

Key Concepts

Integration Techniques

Integration is a fundamental concept in calculus, and there are different techniques to solve these problems. Understanding these techniques can help you tackle various integral problems efficiently.

• **Basic Integration**: The simplest form involves reversing differentiation. Recognizing patterns can make it easier to find the antiderivative.
• **Substitution Method**: This technique is useful when the function is a composition of two functions. By substituting a part of the integrand with a new variable, the integration becomes more manageable.
• **Integration by Parts**: This is based on the product rule for differentiation. It is particularly useful when dealing with products of functions.

By mastering each technique, you will become more comfortable and efficient in solving different types of integrals.

Fundamental Theorem of Calculus

The Fundamental Theorem of Calculus bridges the concept of differentiation and integration together. It has two main parts that make definite integration possible and manageable.

• **First Part**: It states that if a function is continuous over an interval, then the function has an antiderivative on that interval. This implies that integration and differentiation are inverse processes.
• **Second Part**: This part allows us to compute the definite integral of a function using its antiderivative. If F is an antiderivative of f on the interval [a, b], then the integral from a to b of f is given by \( F(b) - F(a) \).

Understanding these concepts is crucial for evaluating definite integrals since these rules provide a straightforward way to find the area under the curve of a function.

Polynomial Integration

Polynomials are among the simplest functions to integrate, and understanding their integration is foundational. Given a polynomial function, the integration process involves increasing the power of each term by one and dividing by the new power.

• **Principle**: If you have a term like \(x^n\), its integral is \(\frac{x^{n+1}}{n+1}\) plus a constant of integration \(C\). This rule applies to all individual terms of a polynomial.
• **Definite Integrals of Polynomials**: Once you have the antiderivative, use the Fundamental Theorem of Calculus to compute its value over specific limits. For a polynomial \(f(x) = x^2\), integrate to get \(\frac{x^3}{3}\), and use specified limits to find the value of the definite integral.

By practicing polynomial integration, you'll gain confidence in integrating more complex functions, as these serve as building blocks for more advanced calculus concepts.