Problem 13 - Applied Calculus

Question

Using the Fundamental Theorem, evaluate the definite integrals in Problems \(1-20\) exactly. $$\int_{0}^{1}\left(y^{2}+y^{4}\right) d y$$

Step-by-Step Solution

Answer

The value of the integral is \( \frac{8}{15} \).

Step 1: Identify the Function and Limits

The given integral is \( \int_{0}^{1}(y^2 + y^4) \, dy \). The function to be integrated is \( f(y) = y^2 + y^4 \), and the limits of integration are from 0 to 1.

Step 2: Find the Antiderivative

To solve the integral, we first need to find the antiderivative of the function \( f(y) = y^2 + y^4 \). The antiderivative of \( y^2 \) is \( \frac{y^3}{3} \), and the antiderivative of \( y^4 \) is \( \frac{y^5}{5} \). Thus, the combined antiderivative is \( F(y) = \frac{y^3}{3} + \frac{y^5}{5} \).

Step 3: Evaluate the Antiderivative at the Upper Limit

Substitute the upper limit of integration (1) into the antiderivative: \( F(1) = \frac{1^3}{3} + \frac{1^5}{5} = \frac{1}{3} + \frac{1}{5} \).

Step 4: Evaluate the Antiderivative at the Lower Limit

Substitute the lower limit of integration (0) into the antiderivative: \( F(0) = \frac{0^3}{3} + \frac{0^5}{5} = 0 \).

Step 5: Apply the Fundamental Theorem of Calculus

According to the Fundamental Theorem of Calculus, the definite integral \( \int_{0}^{1}(y^2 + y^4) \, dy \) is given by \( F(1) - F(0) \). This results in \( \left( \frac{1}{3} + \frac{1}{5} \right) - 0 = \frac{1}{3} + \frac{1}{5} \).

Step 6: Simplify the Result

Compute \( \frac{1}{3} + \frac{1}{5} \) by finding a common denominator. The least common multiple of 3 and 5 is 15. So, \( \frac{1}{3} = \frac{5}{15} \) and \( \frac{1}{5} = \frac{3}{15} \). Add these fractions to get \( \frac{5}{15} + \frac{3}{15} = \frac{8}{15} \).

Key Concepts

Antiderivative

The antiderivative, or the indefinite integral, is a function that reverses the process of differentiation. If you think of differentiation as finding the rate of change of a function, then finding an antiderivative is like tracing back to the original function whose rate of change is given. To find the antiderivative of a function, you look for a function whose derivative matches the given one.

For a simple polynomial term like \( y^n \), the antiderivative can be determined by increasing the power of \( y \) by one, and then dividing by the new exponent. Applying this to \( y^2 + y^4 \):

• The antiderivative of \( y^2 \) is \( \frac{y^3}{3} \) because you increase the exponent from 2 to 3 and divide by 3.
• Similarly, the antiderivative of \( y^4 \) is \( \frac{y^5}{5} \).

Thus, for the function \( y^2 + y^4 \), the antiderivative is \( \frac{y^3}{3} + \frac{y^5}{5} \). This process is integral to understanding how definite integrals are evaluated using the Fundamental Theorem of Calculus.

Definite Integral

A definite integral represents the accumulation of quantities, which can be thought of as the net area under a curve on a graph, between specified boundaries. Unlike the antiderivative, which is a family of functions, the definite integral has specific limits that define its boundaries:

• The integrand, the function to be integrated, is \( y^2 + y^4 \) in this case.
• The limits, 0 and 1 here, define where to start and stop the accumulation.

Using the Fundamental Theorem of Calculus, which states that if \( F \) is an antiderivative of \( f \), then:\[ \int_{a}^{b} f(y) \, dy = F(b) - F(a) \]you can directly compute the definite integral by evaluating the antiderivative at these limits. In our example, evaluating \( F(y) = \frac{y^3}{3} + \frac{y^5}{5} \) at 1 gives us \( \frac{1}{3} + \frac{1}{5} \), and at 0, it gives us 0. The result is the difference \( (\frac{1}{3} + \frac{1}{5}) - 0 = \frac{8}{15} \). Therefore, the definite integral \( \int_{0}^{1} (y^2 + y^4) \, dy \) is \( \frac{8}{15} \).

Integration Limits

Integration limits are the values that define the scope over which a definite integral is calculated. In mathematics, these limits are crucial because they determine the segment of a graph you're focusing on. When specified, as in \( \int_{0}^{1}(y^2 + y^4) \, dy \), the integration limits are typically denoted at top and bottom of the integral sign:

• The number at the top (1 here) is the upper limit.
• The number at the bottom (0 here) is the lower limit.

These limits are more than just placeholders; they guide the evaluation of the integral's antiderivative. The Fundamental Theorem of Calculus utilizes these limits by subtracting the value of the antiderivative at the lower limit from the value at the upper limit, essentially showing how the function grows between these two points. So in this scenario, the built-up area under \( y^2 + y^4 \) from y = 0 to y = 1 was computed to find \( \frac{8}{15} \), reflecting how much 'quantity' exists there.