Problem 58 - Calculus for Biology and Medicine

Question

Given that \(\int_{0}^{a} x^{4} d x=\frac{1}{5} a^{5}\) evaluate the following integrals (a) \(\int_{0}^{2} x^{4} d x\) (b) \(\int_{0}^{1} \frac{x^{4}}{2} d x\) (c) \(\int_{-1}^{1} \frac{x^{4}}{2} d x\) (d) \(\int_{-2}^{0}(x+2)^{4} d x\) (e) \(\int_{-3}^{0}(x+1)^{4} d x\) (f) \(\int_{0}^{2} 2(x-2)^{4} d x\).

Step-by-Step Solution

Answer

(a) \(\frac{32}{5}\); (b) \(\frac{1}{10}\); (c) \(\frac{1}{10}\); (d) \(\frac{32}{5}\); (e) \(-\frac{31}{5}\); (f) \(-\frac{64}{5}\).

Step 1: Evaluate Integral (a)

The problem gives that \( \int_{0}^{a} x^{4} dx = \frac{1}{5}a^{5} \). To find \( \int_{0}^{2} x^{4} dx \), substitute \( a = 2 \) into the given expression. This gives \( \int_{0}^{2} x^{4} dx = \frac{1}{5} \cdot 2^{5} = \frac{1}{5} \cdot 32 = \frac{32}{5} \).

Step 2: Evaluate Integral (b)

Evaluate \( \int_{0}^{1} \frac{x^{4}}{2} dx \). This is the same as \( \frac{1}{2} \int_{0}^{1} x^{4} dx \). From the given, we know \( \int_{0}^{1} x^{4} dx = \frac{1}{5} \cdot 1^{5} = \frac{1}{5} \). Therefore, \( \int_{0}^{1} \frac{x^{4}}{2} dx = \frac{1}{2} \cdot \frac{1}{5} = \frac{1}{10} \).

Step 3: Evaluate Integral (c)

Calculate \( \int_{-1}^{1} \frac{x^{4}}{2} dx \) knowing that the integrand \( \frac{x^{4}}{2} \) is an even function. Therefore, this can be written as \( 2 \times \left( \frac{1}{2} \int_{0}^{1} x^{4} dx \right) = \int_{0}^{1} \frac{x^{4}}{2} dx \). From step (b), we computed this integral to be \( \frac{1}{10} \).

Step 4: Evaluate Integral (d)

Substitute \( u = x+2 \) in the integral \( \int_{-2}^{0}(x+2)^{4} dx \). So, \( du = dx \) and at \( x = -2, u = 0 \) and at \( x = 0, u = 2 \). The integral changes to \( \int_{0}^{2} u^{4} du \), which we calculated as \( \frac{32}{5} \) in step (a).

Step 5: Evaluate Integral (e)

Substitute \( u = x+1 \) in the integral \( \int_{-3}^{0}(x+1)^{4} dx \). Then, \( du = dx \) and at \( x = -3, u = -2 \), and at \( x = 0, u = 1 \). The integral becomes \( \int_{-2}^{1} u^{4} du \). Split it to \( \int_{-2}^{0} u^{4} du + \int_{0}^{1} u^{4} du \), which equals \( -\int_{0}^{2} x^{4} dx + \int_{0}^{1} x^{4} dx = -\frac{32}{5} + \frac{1}{5} = -\frac{31}{5} \).

Step 6: Evaluate Integral (f)

Perform substitution \( u = x-2 \) for \( \int_{0}^{2} 2(x-2)^{4} dx \). Here, \( du = dx \), and at \( x = 0, u = -2 \), and at \( x = 2, u = 0 \). The integral becomes \( 2 \int_{-2}^{0} u^{4} du \), which equals \( 2(-\int_{0}^{2} x^{4} dx) = -2 \times \frac{32}{5} = -\frac{64}{5} \).

Key Concepts

Definite Integrals

A definite integral represents the signed area under a curve on a particular interval on the x-axis. This means it calculates the accumulation of quantities such as area, volume, or other values derived from a function over the specified interval. For instance, the expression \[ \int_{a}^{b} f(x) \, dx \] outputs the area under the curve of \( f(x) \) from \( x = a \) to \( x = b \).
It's important to note that a definite integral is different from an indefinite integral, which represents a family of functions or the antiderivative of a function. With definite integrals, the limits \( a \) and \( b \) are crucial as they define the region to be evaluated.

• For example, in the solution for \( \int_{0}^{2} x^{4} \, dx \), by substituting \( a = 2 \) in the formula \( \int_{0}^{a} x^{4} \, dx = \frac{1}{5} a^{5} \), we found \( \int_{0}^{2} x^{4} \, dx = \frac{32}{5} \). This represents the volume or accumulation between \( 0 \) and \( 2 \) along the function \( x^4 \).

Understanding definite integrals underpins much of integral calculus and their applications in physics, engineering, and quantitative fields.

Function Substitution

Function substitution is a method in calculus where we replace variables in a given problem to simplify an integral or differential equation. In substitution, we reset the limits of integration or find a new approach to simplify the expression being integrated.
This involves selecting a substitution \( u(x) \) to replace \( x \) and calculating the derivative \( du = g'(x) \cdot dx \), altering the integral into a simpler form that is easier to evaluate.

• In evaluating \( \int_{-2}^{0} (x+2)^{4} \, dx \), we substituted \( u = x + 2 \), changing limits from \( x = -2 \) to \( x = 0 \) into \( u = 0 \) to \( u = 2 \).
• After substitution, our problem becomes \( \int_{0}^{2} u^{4} \, du \), which we've encountered before and calculated easily using known results.

Function substitution is especially valuable in making complex integrations manageable, turning intimidating definite integrals into straightforward calculations.

Even Function Property

The even function property is particularly useful in definite integrals when dealing with symmetry around the y-axis. An even function \( f(x) \) satisfies \( f(-x) = f(x) \) for all values of \( x \). The symmetry allows us to simplify integrals spanning from \( -a \) to \( a \).Consider the integral \[ \int_{-a}^{a} f(x) \, dx \].
If \( f(x) \) is even, this integral can be calculated as \[ 2 \int_{0}^{a} f(x) \, dx \] because of the mirror-like symmetry.

• For example, in evaluating \( \int_{-1}^{1} \frac{x^{4}}{2} \, dx \), the integrand \( \frac{x^{4}}{2} \) is an even function. This means that the integral from \( -1 \) to \( 1 \) simplifies to \( 2 \int_{0}^{1} \frac{x^{4}}{2} \, dx \), effectively making our calculation easier.

Recognizing and leveraging even function properties drastically reduce the complexity of calculations and highlight integral calculus's power in exploring and exploiting function symmetries.