Question

Variations on the substitution method Evaluate the following integrals. $$\int x(x+10{)}^{9}dx$$

Short answer

The method used to solve this problem was the substitution method (also known as the change of variables method or 'u-substitution'). The chosen substitution for this problem was $$u=x+10$$.

Step-by-step solution

Identify the substitution

The function given here is $$x(x+10{)}^9$$. It can be broken down into a product of two functions, where one function is part of the derivative of the other function. In this case, let $$u=x+10$$. Then, the derivative of $$u$$ is $$\frac{du}{dx}=1$$, which is a simple constant.

Replace dx and substitute u

We now want to replace $$dx$$ in the original integral expression. To do so, we have $$du=dx$$ since the derivative of $$u$$ with respect to $$x$$ is a constant $$(1)$$. Our integral becomes $$\int x(u{)}^{9}du.$$Now notice that we still have an $$x$$ in this expression.

Solving for x and substituting

We can solve for $$x$$ from our substitution, $$u=x+10$$. That gives us $$x=u-10$$. Now, replace $$x$$ with $$(u-10)$$ in the integral: $$\int (u-10)(u{)}^{9}du$$Now the integral is prepared to be solved.

Expanding the integral

Expand the integral to be able to find the antiderivative: $$\int (u^{10}-10u^9)du$$

Finding the antiderivative

Now, find the antiderivative of the expanded integral: $$\int (u^{10}-10u^9)du=\frac{1}{11}{u}^{11}-u^{10}+C$$

Substituting back for x

Now we substitute back for $$x$$, replacing $$u$$ with $$(x+10)$$ to get the final answer: $$\frac{1}{11}(x+10{)}^{11}-(x+10{)}^{10}+C$$ So, the evaluated integral is: $$\int x(x+10{)}^{9}dx=\frac{1}{11}(x+10{)}^{11}-(x+10{)}^{10}+C$$

Key concepts

integration

Integration is a fundamental concept in calculus that is used to find the area under a curve or to accumulate quantities. When we integrate a function, we essentially reverse the process of differentiation. In other words, integration is the process of finding a function whose derivative is the given function. Integrals can be classified into two major types: definite integrals and indefinite integrals.
When we perform a definite integral, we are finding the exact area under a curve between two specific points. On the other hand, an indefinite integral, which is what we typically deal with in problems using the substitution method, gives us a family of functions known as antiderivatives.

• Integration is crucial for solving differential equations, which describe how quantities change.
• In the context of our exercise, integration allows us to evaluate complex expressions like $\int x(x+10{)}^{9}dx$.

The substitution method is a popular technique used in integration to simplify expressions. In this technique, we essentially substitute part of the integral with a single variable, allowing us to convert a complex integral into a simpler, more manageable one. This is particularly useful when the integral involves compositions of functions, where one function is the derivative of another. Substitution makes the process of finding integrals more manageable. By substituting $u=x+10$, the integral in our exercise becomes easier to handle.

antiderivatives

An antiderivative, intuitively, is a function that, when differentiated, yields the original function. When we talk about finding the integral of a function, we mean finding its antiderivative. Every function has an infinite number of antiderivatives, differing only by a constant. This is why we often write the general solution of an indefinite integral with a "+ C" to account for all possible antiderivatives. This "C" is known as the constant of integration.
Antiderivatives are crucial in problem-solving as they help us reverse the process of differentiation. This can be particularly useful in physics when we want to determine position functions from velocity functions.
In the context of this exercise, by finding the antiderivative of $u^{10}-10u^9$, we are determining the function that, when differentiated, gives us the original polynomial. Thus, the expression becomes $\frac{1}{11}{u}^{11}-u^{10}+C$, which is then back-substituted to replace $u$ with $x+10$.

• Antiderivatives are also called indefinite integrals.
• They play a key role in solving problems involving accumulation or areas under curves.

integral calculus

Integral calculus is the branch of calculus that focuses on functions, their integrals, and antiderivatives. It complements differential calculus, which is concerned with rates of change and slopes of curves.
While differentiation deals with breaking down functions into their constituent rates of change, integral calculus is about collecting them back into a full picture. This often involves reversing the differentiation process to recover functions from their rates of change. The two primary operations in integral calculus are finding definite and indefinite integrals.

• Integral calculus helps find solutions to real-world problems involving areas, volumes, and other accumulations.
• It's foundational in fields such as physics, engineering, and economics where understanding quantities and their changes is essential.

In our exercise, integral calculus was applied through substitution to simplify the integrand. After substituting $u=x+10$, we utilized integral calculus concepts to expand the expression and find the antiderivative. By back-substituting $u$ with $x+10$, we returned to the variable of the original problem, providing a complete solution that exemplifies the power of integral calculus in solving complex integrals.