Problem 18 - Applied Calculus

Question

Find the integrals in problems. Check your answers by differentiation. $$ \int y^{2}(1+y)^{2} d y $$

Step-by-Step Solution

Answer

The integral is \( \frac{y^3}{3} + \frac{y^4}{2} + \frac{y^5}{5} + C \).

Step 1: Expand the Integrand

The integrand is given as \( y^2 (1+y)^2 \). First, expand \( (1+y)^2 \) to \( 1 + 2y + y^2 \). Then distribute \( y^2 \) across this expression to get \( y^2 + 2y^3 + y^4 \).

Step 2: Integrate Each Term

Now that the integrand is expanded to \( y^2 + 2y^3 + y^4 \), integrate each term separately:\[ \int y^2 \, dy = \frac{y^3}{3}, \]\[ \int 2y^3 \, dy = \frac{2y^4}{4} = \frac{y^4}{2}, \]\[ \int y^4 \, dy = \frac{y^5}{5}. \]

Step 3: Write the General Solution

Combine the integrals calculated to form the general solution:\[ \int y^2 (1+y)^2 \, dy = \frac{y^3}{3} + \frac{y^4}{2} + \frac{y^5}{5} + C, \]where \(C\) is the constant of integration.

Step 4: Differentiate the Solution to Verify

Differentiate the function \( \frac{y^3}{3} + \frac{y^4}{2} + \frac{y^5}{5} + C \) with respect to \(y\):\( \frac{d}{dy} \left( \frac{y^3}{3} \right) = y^2 \),\( \frac{d}{dy} \left( \frac{y^4}{2} \right) = 2y^3 \),\( \frac{d}{dy} \left( \frac{y^5}{5} \right) = y^4 \),and the derivative of \(C\) is zero. Summing these gives the original integrand \(y^2 + 2y^3 + y^4\), confirming our answer is correct.

Key Concepts

Differentiation

Differentiation is the mathematical process of finding the derivative of a function. A derivative represents how a function changes as its input changes. In essence, it measures the rate at which one quantity changes with respect to another. Differentiation can be thought of as a tool that

• Shows how functions grow or decay
• Provides insights into the behavior of functions
• Helps verify integrals

In the context of this exercise, we used differentiation to check our integral result. After integrating the expanded expression, we ended up with a function. By differentiating this function, we expected to obtain the original integrand. This process confirmed the accuracy of our integration. Differentiating indeed retraces our steps back to the initial expression, underpinning the fundamental theorem of calculus which lovingly ties differentiation and integration together.

Integrand Expansion

Integrand expansion is a technique used to simplify an integral by expanding complex parts of it into simpler components. This method is particularly useful when an integrand is given in a compound form, such as a product of two expressions. By expanding these expressions, integration becomes more manageable.
In the exercise we addressed, we started with the integrand \(y^2 (1+y)^2\). The first step taken was to expand the \((1+y)^2\) expression, which simplifies into \(1 + 2y + y^2\). Then we distributed \(y^2\) across this new expression. The result was \(y^2 + 2y^3 + y^4\), a format that's straightforward to integrate.

When you face a complex integrand, consider expansion as your first step. It not only makes the integration more intuitive but ensures you can apply standard techniques like the power rule with ease.

Power Rule for Integration

The power rule for integration is a cornerstone method in calculus. It's widely used for integrating simple power functions of the form \(x^n\). The rule states that to integrate \(x^n\), you increase the exponent by 1 and divide by the new exponent:
\[ \int x^n \, dx = \frac{x^{n+1}}{n+1} + C \]
where \(C\) is the constant of integration. This rule is simple yet powerful, applicable directly after expansion and simplification of the integrand.

• For \(y^2\), using the power rule gives \(\frac{y^3}{3}\)
• For \(2y^3\), you compute \(\int 2y^3 \, dy = \frac{y^4}{2}\)
• For \(y^4\), the result is \(\frac{y^5}{5}\)

By applying the power rule to each term separately, we obtain the integrated expression. This systematic approach simplifies the computation and is central to solving a wide range of integrals effectively.