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Chapter 5: Problem 39

Find each indefinite integral. \(\int(t+1)^{3} d t\)

Short Answer

Expert verified
The integral is \( \frac{(t+1)^4}{4} + C \).

Step by step solution

01

Identify the Function to Integrate

Recognize that the given integral is \( \int (t+1)^3 \, dt \). We need to find its antiderivative.
02

Choose an Appropriate Substitution

Use the substitution method. Let \( u = t+1 \), hence \( du = dt \). The integral becomes \( \int u^3 \, du \).
03

Integrate the New Function

Integrate \( \int u^3 \, du \) using the power rule for integration: \( \int u^n \, du = \frac{u^{n+1}}{n+1} + C \), where \( C \) is the constant of integration. Apply this to get \( \frac{u^4}{4} + C \).
04

Substitute Back to Original Variable

Substitute \( u = t+1 \) back into the equation to express the antiderivative in terms of \( t \). The antiderivative is \( \frac{(t+1)^4}{4} + C \).
05

State the Final Answer

The indefinite integral of \( \int (t+1)^3 \, dt \) is \( \frac{(t+1)^4}{4} + C \). This is the general solution with the arbitrary constant \( C \).

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Key Concepts

These are the key concepts you need to understand to accurately answer the question.

Substitution Method
The substitution method is like a handy toolbox for solving integrals, especially when faced with composite functions. Think of it as changing the perspective to make the problem simpler. In this method, we replace a part of the integrand (the expression you want to integrate) with a new variable, usually one that simplifies the expression.

  • Start by identifying a substitution that simplifies the integrand. In our exercise, we set u equal to t + 1.
  • Next, express the differential in terms of the new variable. In this case, since u = t + 1, we have du = dt. This simple relation helps transform the integral into something more manageable.
By substituting, our integral \( \int (t+1)^3 \, dt \) becomes \( \int u^3 \, du \). This transformation is crucial. It makes the integral look straightforward to tackle with basic integration rules. After integrating in terms of u, we switch back to the original variable t to find our solution.
Power Rule for Integration
The power rule for integration is an essential tool, especially when dealing with polynomials. It allows us to easily find the integral of power functions.

Here鈥檚 how it works:
  • When you see an integral of the form \( \int x^n \, dx \), you use the power rule, which states: \( \int x^n \, dx = \frac{x^{n+1}}{n+1} + C \), provided \( n eq -1 \).
  • The constant \( C \) is added to represent the constant of integration, important in indefinite integrals. It accounts for any constant function that could differentiate to zero.
In our exercise, after substituting, we use this rule to integrate \( \int u^3 \, du \). Here, n is 3, so we increase the exponent by one, making it 4, and divide by the new exponent. Hence, \( \frac{u^4}{4} + C \).This application simplifies polynomial integration, especially after using substitution to reframe the problem.
Antiderivative
An antiderivative is like the reverse of differentiation. While differentiation tells us the rate of change, finding an antiderivative gives us the original function before it was differentiated.

With indefinite integrals, we're looking for a family of functions whose derivative matches the given integrand.
  • In symbols, if \( F'(x) = f(x) \), then \( F(x) \) is an antiderivative of \( f(x) \).
  • The notation \( \int f(x) \, dx \) represents finding such an antiderivative, denoted also as \( F(x) + C \), with \( C \) being an arbitrary constant.
So, for our integral \( \int (t+1)^3 \, dt \), we find the antiderivative as \( \frac{(t+1)^4}{4} + C \). This expression signifies all possible functions that can differentiate back to the original integrand, \( (t+1)^3 \).

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Most popular questions from this chapter

\(85-94 .\) The substitution method can be used to find integrals that do not fit our formulas. For example, observe how we find the following integral using the substitution \(u=x+4\) which implies that \(x=u-4\) and so \(d x=d u\). $$ \begin{aligned} \int(x-2)(x+4)^{8} d x &=\int(u-4-2) u^{8} d u \\ &=\int(u-6) u^{8} d u \\ &=\int\left(u^{9}-6 u^{8}\right) d u \\ &=\frac{1}{10} u^{10}-\frac{2}{3} u^{9}+C \\ &=\frac{1}{10}(x+4)^{10}-\frac{2}{3}(x+4)^{9}+C \end{aligned} $$ It is often best to choose \(u\) to be the quantity that is raised to a power. The following integrals may be found as explained on the left (as well as by the methods of Section 6.1). $$ \int(x+1)(x-5)^{4} d x $$

The substitution method can be used to find integrals that do not fit our formulas. For example, observe how we find the following integral using the substitution \(u=x+4\) which implies that \(x=u-4\) and so \(d x=d u\). $$ \begin{aligned} \int(x-2)(x+4)^{8} d x &=\int(u-4-2) u^{8} d u \\ &=\int(u-6) u^{8} d u \\ &=\int\left(u^{9}-6 u^{8}\right) d u \\ &=\frac{1}{10} u^{10}-\frac{2}{3} u^{9}+C \\ &=\frac{1}{10}(x+4)^{10}-\frac{2}{3}(x+4)^{9}+C \end{aligned} $$ It is often best to choose \(u\) to be the quantity that is raised to a power. The following integrals may be found as explained on the left (as well as by the methods of Section 6.1). $$ \int \frac{x}{\sqrt[3]{x+1}} d x $$

The substitution method can be used to find integrals that do not fit our formulas. For example, observe how we find the following integral using the substitution \(u=x+4\) which implies that \(x=u-4\) and so \(d x=d u\). $$ \begin{aligned} \int(x-2)(x+4)^{8} d x &=\int(u-4-2) u^{8} d u \\ &=\int(u-6) u^{8} d u \\ &=\int\left(u^{9}-6 u^{8}\right) d u \\ &=\frac{1}{10} u^{10}-\frac{2}{3} u^{9}+C \\ &=\frac{1}{10}(x+4)^{10}-\frac{2}{3}(x+4)^{9}+C \end{aligned} $$ It is often best to choose \(u\) to be the quantity that is raised to a power. The following integrals may be found as explained on the left (as well as by the methods of Section 6.1). $$ \int \frac{x-4}{x-5} d x $$

Find the area between the curve \(y=x e^{x^{2}}\) and the \(x\) -axis from \(x=1\) to \(x=3\) (Leave the answer in its exact form.)

In 2000 the birthrate in Africa increased from \(17 e^{0.02 t}\) million births per year to \(22 e^{0.02 t}\) million births per year, where \(t\) is the number of years since 2000 . Find the total increase in population that will result from this higher birthrate between \(2000(t=0)\) and \(2050(t=50)\).
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