Problem 12 - Applied Calculus

Question

Using the Fundamental Theorem, evaluate the definite integrals in problem exactly. $$ \int_{0}^{1}\left(y^{2}+y^{4}\right) d y $$

Step-by-Step Solution

Answer

The definite integral evaluates to \( \frac{8}{15} \).

Step 1: Identify the Function to Integrate

Examine the given integral \( \int_{0}^{1}(y^2 + y^4) \, dy \). Here, the function to integrate is \( f(y) = y^2 + y^4 \).

Step 2: Integrate the Function

Find the indefinite integral of \( f(y) = y^2 + y^4 \). The integral of \( y^2 \) is \( \frac{y^3}{3} \), and the integral of \( y^4 \) is \( \frac{y^5}{5} \). Hence, the indefinite integral is \( \int (y^2 + y^4) \, dy = \frac{y^3}{3} + \frac{y^5}{5} + C \), where \( C \) is the constant of integration.

Step 3: Apply the Fundamental Theorem of Calculus

Using the Fundamental Theorem of Calculus, evaluate \( \left[ \frac{y^3}{3} + \frac{y^5}{5} \right]_{0}^{1} \). This means substituting the upper limit and lower limit into the integral expression and finding their difference.

Step 4: Compute the Integral at the Upper Limit

Substitute \( y = 1 \) into \( \frac{y^3}{3} + \frac{y^5}{5} \) to get \( \frac{1^3}{3} + \frac{1^5}{5} = \frac{1}{3} + \frac{1}{5} \).

Step 5: Compute the Integral at the Lower Limit

Substitute \( y = 0 \) into \( \frac{y^3}{3} + \frac{y^5}{5} \) to get \( \frac{0^3}{3} + \frac{0^5}{5} = 0 \).

Step 6: Calculate the Definite Integral

Subtract the lower limit value from the upper limit value: \( \left( \frac{1}{3} + \frac{1}{5} \right) - 0 = \frac{1}{3} + \frac{1}{5} \).

Step 7: Simplify the Result

Convert \( \frac{1}{3} + \frac{1}{5} \) to a common denominator for easy addition. The least common denominator is 15, so \( \frac{1}{3} = \frac{5}{15} \) and \( \frac{1}{5} = \frac{3}{15} \). Add these to get \( \frac{5}{15} + \frac{3}{15} = \frac{8}{15} \).

Key Concepts

Definite Integrals

Definite integrals are an essential concept in calculus, allowing us to calculate the area under a curve over a specified interval.
Essentially, a definite integral takes a function and computes the accumulation of its values over a range.
In symbol terms, for the function \( f(x) \) over the interval \([a, b]\), the definite integral is denoted as \( \int_{a}^{b} f(x) \, dx \).

• The definite integral evaluates to a number rather than a function because it reflects the total accumulation of the area under the curve from \(a\) to \(b\).


• Definite integrals can be solved by using the Fundamental Theorem of Calculus, which bridges the concept of integration with differentiation by allowing us to evaluate integrals using antiderivatives.

In the context of the exercise, we needed to evaluate \( \int_{0}^{1}(y^2 + y^4) \, dy \). Using the steps provided, a value (\( \frac{8}{15} \)) was calculated, representing the area under the function \(y^2 + y^4\) from \(y=0\) to \(y=1\). This result is an example of how definite integrals give concrete, numerical results that have meaningful interpretations in real-world applications.

Indefinite Integrals

Indefinite integrals are closely related to definite integrals, yet serve a different purpose.
Whereas definite integrals result in a specific number, indefinite integrals represent a family of functions.
Mathematically, for a function \( f(x) \), the indefinite integral is written as \( \int f(x) \, dx \).

• The result of an indefinite integral is an antiderivative of the function, represented with a constant of integration \( C \) (e.g., \( F(x) + C \)). This constant represents an infinite number of functions that differ by a constant value.


• Indefinite integrals are useful for understanding the general behavior and shape of antiderivatives without focusing on specific numerical values.

In the exercise, when finding the indefinite integral of \( y^2 + y^4 \), we calculated \( \int (y^2 + y^4) \, dy = \frac{y^3}{3} + \frac{y^5}{5} + C \). The term \(+C\) is crucial because it indicates that the solution is valid for any vertical shift of the function, hence representing all potential antiderivatives.

Integration Techniques

Understanding integration techniques is vital as they allow us to solve both definite and indefinite integrals.
Various techniques help handle the wide range of functions encountered in calculus.
Some fundamental techniques include:

• Basic Integration Rules: These include the power rule, constant rule, and sum rule, which simplify handling straightforward polynomial and simple functions.


• Integration by Substitution: Useful for integrals of composite functions, this technique reduces a complex integral to a simpler form by changing variables.


• Integration by Parts: This technique uses the product rule of differentiation in reverse, breaking down integrals of products of functions.

In the exercise discussed, we primarily used basic integration rules directly. For instance, the power rule was applied to find the integral of \( y^2 \) and \( y^4 \), arriving at \( \frac{y^3}{3} \) and \( \frac{y^5}{5} \) respectively. Such foundational techniques are the building blocks for tackling more complex integrals, making them essential for understanding calculus.