Problem 9 - Biocalculus Calculus for the Life Sciences

Question

Evaluate the integral. \(\int_{1}^{2}(1+2 y)^{2} d y\)

Step-by-Step Solution

Answer

The integral evaluates to \(\frac{49}{3}\).

Step 1: Expand the Integrand

We start by expanding the integrand \((1 + 2y)^2\). Use binomial expansion: \((a + b)^2 = a^2 + 2ab + b^2\). Letting \(a = 1\) and \(b = 2y\), we have: \((1+2y)^2 = 1^2 + 2 \times 1 \times 2y + (2y)^2 = 1 + 4y + 4y^2\).

Step 2: Set Up the Integral with Expanded Form

Substitute the expanded expression back into the integral, we have:\[ \int_{1}^{2} (1 + 4y + 4y^2) \, dy \]

Step 3: Integrate Term by Term

Now, integrate each term separately:- The integral of 1 with respect to y is: \( \int 1 \, dy = y \).- The integral of \(4y\) is: \( \int 4y \, dy = 4 \cdot \frac{y^2}{2} = 2y^2 \).- The integral of \(4y^2\) is: \( \int 4y^2 \, dy = 4 \cdot \frac{y^3}{3} = \frac{4y^3}{3} \).

Step 4: Combine the Antiderivatives

Combine all the antiderivatives calculated:\[ y + 2y^2 + \frac{4y^3}{3} \]

Step 5: Evaluate the Definite Integral

Use the fundamental theorem of calculus to evaluate from 1 to 2:- First, substitute 2: \( 2 + 2 \times 2^2 + \frac{4 \times 2^3}{3} = 2 + 8 + \frac{32}{3} = 10 + \frac{32}{3} = \frac{30}{3} + \frac{32}{3} = \frac{62}{3} \).- Next, substitute 1: \( 1 + 2 \times 1^2 + \frac{4 \times 1^3}{3} = 1 + 2 + \frac{4}{3} = 3 + \frac{4}{3} = \frac{9}{3} + \frac{4}{3} = \frac{13}{3} \).Finally, subtract: \( \frac{62}{3} - \frac{13}{3} = \frac{49}{3} \).

Step 6: Final Result

State the result of the evaluated definite integral: \( \int_{1}^{2} (1 + 2y)^2 \, dy = \frac{49}{3} \).

Key Concepts

Binomial Expansion

When you encounter an expression like \((1 + 2y)^2\), the binomial expansion is a handy tool. It helps simplify expressions raised to a power. For a binomial expression \((a + b)^2\), the expansion formula is \(a^2 + 2ab + b^2\). This formula allows us to break down the expression into simpler terms.

In the given exercise, we have \(a = 1\) and \(b = 2y\). Using the formula, we expand \((1 + 2y)^2\) into \(1^2 + 2 \times 1 \times 2y + (2y)^2\). This results in \(1 + 4y + 4y^2\).

By expanding expressions, it becomes easier to integrate each term individually. This step is crucial for managing complex polynomials and simplifying calculation tasks in calculus. Remembering the binomial expansion formula can greatly aid in solving integral and other algebraic problems effectively.

Antiderivative

Integrating involves finding the antiderivative, which is essentially the reverse of taking a derivative. To solve an integral like \(\int (1 + 4y + 4y^2) \, dy\), you need to find the antiderivative for each term separately.

• The antiderivative of a constant \(1\) is simply \(y\), because the derivative of \(y\) is \(1\).
• For the term \(4y\), the antiderivative is \(2y^2\). This follows the power rule, where you increase the exponent by one and divide by the new exponent.
• The term \(4y^2\) involves slightly more calculation, leading to \(\frac{4y^3}{3}\). Again, using the power rule helps: add 1 to the power and divide by the new exponent.

Remember, the power rule is \(\int x^n \, dx = \frac{x^{n+1}}{n+1}\) for any non-zero \(n\). After finding each antiderivative, they are combined to form the function needed for the next step, definite integration.

Fundamental Theorem of Calculus

The Fundamental Theorem of Calculus forms a bridge between antiderivatives and definite integrals. It's composed of two parts: finding the antiderivative and evaluating the definite integral. This theorem states that if you have a continuous function, the integral from \(a\) to \(b\) of the function is found by evaluating the antiderivative at \(b\) minus the antiderivative at \(a\).

In this context, after finding the combined antiderivative function \(y + 2y^2 + \frac{4y^3}{3}\), you then evaluate it at the limits of integration, which are 2 and 1.

• First, substitute the upper limit (2) into the antiderivative to get \(y(2) = 2 + 8 + \frac{32}{3} = \frac{62}{3}\).
• Next, substitute the lower limit (1) into the antiderivative to get \(y(1) = 1 + 2 + \frac{4}{3} = \frac{13}{3}\).
• The final calculation \(\frac{62}{3} - \frac{13}{3} = \frac{49}{3}\) gives the area under the curve from 1 to 2.

This crucial step showcases the power of the Fundamental Theorem of Calculus in connecting limits of integration with antiderivatives, ensuring you can solve any definite integral once the antiderivative is known.