Question

Variations on the substitution method Evaluate the following integrals. $$\int \frac{y^2}{(y+1{)}^4}dy$$

Short answer

Question: Evaluate the integral of $\frac{y^2}{(y+1{)}^4}$ with respect to y. Answer: The integral of $\frac{y^2}{(y+1{)}^4}$ with respect to y is $-\frac{1}{y+1}-\frac{1}{(y+1{)}^2}-\frac{1}{3(y+1{)}^3}+C$, where C is the constant of integration.

Step-by-step solution

Let $u=y+1$. Now we need to differentiate $u$ to find $du$ in terms of $dy$. #Step 2: Differentiate the substitution variable#

Differentiate $u$ with respect to $y$ to get $du/dy$. So, we have: $$\frac{du}{dy}=1$$ Now, multiply both sides by $dy$: $$du=dy$$. #Step 3: Rewrite the integral in terms of u#

Now we can rewrite the integral in terms of $u$. The integral becomes: $$\int \frac{(u-1{)}^2}{u^4}du$$ #Step 4: Simplify the integrand#

Expand $(u-1{)}^2$ and simplify the expression: $$\int \frac{u^2-2u+1}{u^4}du$$ Now rewrite the integrand as the sum of three separate fractions: $$\int (\frac{1}{u^2}-\frac{2}{u^3}+\frac{1}{u^4})du$$ #Step 5: Integrate#

Integrate each term separately: $$\int \frac{1}{u^2}du-\int \frac{2}{u^3}du+\int \frac{1}{u^4}du$$ The integrals are: $$\int u^{-2}du=-u^{-1}+C_1=-\frac{1}{u}+C_1$$ $$\int 2u^{-3}du=-u^{-2}+C_2=-\frac{1}{u^2}+C_2$$ $$\int u^{-4}du=-\frac{1}{3}{u}^{-3}+C_3=\frac{-1}{3u^3}+C_3$$ Add these three integrals together (including their constants): $$-\frac{1}{u}-\frac{1}{u^2}-\frac{1}{3u^3}+C$$ where $C=C_1+C_2+C_3$. #Step 6: Substitute back the original variable#

Now substitute back $y+1$ for $u$. The final answer is: $$\int \frac{y^2}{(y+1{)}^4}dy=-\frac{1}{y+1}-\frac{1}{(y+1{)}^2}-\frac{1}{3(y+1{)}^3}+C$$

Key concepts

Indefinite Integral

The concept of the indefinite integral is fundamental in calculus. It represents the collection of all antiderivatives of a function. When we write the integral of a function without specifying the limits of integration, such as $\text{[}\int f(x)dx\text{]}$, we are dealing with an indefinite integral. This operation essentially reverses differentiation, and the result is often expressed with a '+ C' at the end, where 'C' denotes the constant of integration. This constant is included because differentiation of a constant yields zero, so when taking the antiderivative, there are actually infinitely many functions that differ by a constant value.

The solution to an indefinite integral is not a single function, but a family of functions. In the original exercise, for example, after integrating, we add '+ C' to account for all possible antiderivatives of the given function.

U-Substitution

The technique of u-substitution is a method used to simplify the integration process. It involves a change of variable to transform an integral into a simpler form that is easier to evaluate. To perform u-substitution, we choose a new variable 'u' which is a function of 'x', typically in a way that makes the integral look simpler or more familiar. We then express 'dx' in terms of 'du' and replace all instances of 'x' in the integral with expressions in terms of 'u'.

After the substitution is made and the integral is evaluated with respect to 'u', it is important to substitute back to the original variable to get the result in terms of the original variable, as seen in step 6 of the solution where 'u' is replaced with 'y+1' to obtain the final answer in terms of 'y'. U-substitution is not only a powerful integration technique but also an excellent example of how a clever change of perspective can simplify a problem.

Integration Techniques

In calculus, integration techniques are various strategies used to evaluate integrals. Integration can be challenging because it is often not straightforward to find the antiderivative of a complex function. There are several methods to tackle difficult integrals, including u-substitution, integration by parts, trigonometric substitution, partial fractions, and integration of rational functions, to name a few.

Each method has its own application depending on the shape and complexity of the function one is trying to integrate. The choice of technique can greatly affect the difficulty of the calculation. For instance, the substitution method applied to the given exercise turns a daunting integrand into separate, more manageable terms. It's important for students to become familiar with these techniques and know when to apply them to solve integral problems effectively.

Calculus

At its core, calculus is the study of change and motion. It is a mathematical discipline that provides tools for understanding the ways in which functions change, the relationships between varying quantities, and how to accumulate quantities, such as areas under curves or volumes. Calculus is divided into two main branches: differential calculus, which concerns the concept of the derivative, and integral calculus, which deals with integrals.

Calculus is crucial in many fields such as physics, engineering, economics, statistics, and even medicine. Many of the phenomena in the universe can be described using calculus, making it an essential component of modern science and mathematics education. Understanding the principles of calculus, such as integration and the techniques to perform it, enables students to solve complex problems and gain deeper insights into the functions and models that describe the world around us.