Problem 54
Question
Use integration by substitution and the Fundamental Theorem to evaluate the definite integrals. $$\int_{0}^{3} \frac{1}{\sqrt{t+1}} d t$$
Step-by-Step Solution
Verified Answer
The value of the integral is 2.
1Step 1: Identify the substitution
We will use the substitution method. Let us substitute \( u = t + 1 \). Then, differentiate both sides with respect to \( t \) to obtain \( du = dt \).
2Step 2: Change the limits of integration
Substitute the limits of the integral. When \( t = 0 \), \( u = 0 + 1 = 1 \). When \( t = 3 \), \( u = 3 + 1 = 4 \). Thus, the integral becomes \( \int_{1}^{4} \frac{1}{\sqrt{u}} \, du \).
3Step 3: Simplify and integrate
The integral \( \int \frac{1}{\sqrt{u}} \, du \) can be rewritten using exponent rules. It is equivalent to \( \int u^{-1/2} \, du \). The antiderivative is \( 2u^{1/2} + C \).
4Step 4: Evaluate the definite integral
Apply the Fundamental Theorem of Calculus to evaluate the integral from 1 to 4. The expression becomes \( 2u^{1/2} \Bigg|_{1}^{4} \). Calculate this as follows: \( 2(4^{1/2}) - 2(1^{1/2}) = 2(2) - 2(1) = 4 - 2 = 2 \).
Key Concepts
Definite IntegralsFundamental Theorem of CalculusAntiderivatives
Definite Integrals
A definite integral calculates the area under a curve between two specified points, often referred to as the limits of integration. This can be visualized as the total "net area" between the curve and the x-axis from the lower limit to the upper limit. In definite integrals, the limits of integration are important as they specify where we start and stop the integration process.
When we evaluate the definite integral \( \int_{0}^{3} \frac{1}{\sqrt{t+1}} \, dt \), we use the limits 0 and 3. After substitution in the original problem, these limits change to 1 and 4, which specifies the new range for our variable \( u \). This substitution process transforms the original problem, helping simplify the calculation of the area under the curve defined by the function.
Definite integrals are unique because they produce a numerical result, unlike indefinite integrals, which yield a function plus a constant. When working with definite integrals, the result represents an exact value, like the accumulated quantity or total change over a specified interval. This makes definite integrals particularly useful in physics and engineering to determine things like distance, area, volume, and more.
When we evaluate the definite integral \( \int_{0}^{3} \frac{1}{\sqrt{t+1}} \, dt \), we use the limits 0 and 3. After substitution in the original problem, these limits change to 1 and 4, which specifies the new range for our variable \( u \). This substitution process transforms the original problem, helping simplify the calculation of the area under the curve defined by the function.
Definite integrals are unique because they produce a numerical result, unlike indefinite integrals, which yield a function plus a constant. When working with definite integrals, the result represents an exact value, like the accumulated quantity or total change over a specified interval. This makes definite integrals particularly useful in physics and engineering to determine things like distance, area, volume, and more.
Fundamental Theorem of Calculus
The Fundamental Theorem of Calculus bridges the concepts of differentiation and integration, which are two core processes in calculus. In essence, it provides a way to evaluate a definite integral using an antiderivative. Here is how it works:
- The first part of the theorem tells us that the integral of a function \( f(x) \) from \( a \) to \( b \) can be found using its antiderivative \( F(x) \).- Specifically, if we know an antiderivative \( F(x) \) of \( f(x) \), then the definite integral of \( f(x) \) from \( a \) to \( b \) is \( F(b) - F(a) \).
In the problem \( \int_{1}^{4} \frac{1}{\sqrt{u}} \, du \), the Fundamental Theorem of Calculus allows us to take the antiderivative \( 2u^{1/2} \) and evaluate it at the upper limit, 4, and the lower limit, 1. This subtraction \( F(4) - F(1) \) gives us the exact area under the curve from 1 to 4 for the function \( \frac{1}{\sqrt{u}} \).
This theorem is powerful because it simplifies calculating integrals, transforming what might be a difficult process into a straightforward task involving antiderivatives.
- The first part of the theorem tells us that the integral of a function \( f(x) \) from \( a \) to \( b \) can be found using its antiderivative \( F(x) \).- Specifically, if we know an antiderivative \( F(x) \) of \( f(x) \), then the definite integral of \( f(x) \) from \( a \) to \( b \) is \( F(b) - F(a) \).
In the problem \( \int_{1}^{4} \frac{1}{\sqrt{u}} \, du \), the Fundamental Theorem of Calculus allows us to take the antiderivative \( 2u^{1/2} \) and evaluate it at the upper limit, 4, and the lower limit, 1. This subtraction \( F(4) - F(1) \) gives us the exact area under the curve from 1 to 4 for the function \( \frac{1}{\sqrt{u}} \).
This theorem is powerful because it simplifies calculating integrals, transforming what might be a difficult process into a straightforward task involving antiderivatives.
Antiderivatives
An antiderivative is a function whose derivative is the original function you are working with. Finding an antiderivative is the first step to solving an integral problem. Antiderivatives provide the "reverse" of differentiation, essentially undoing the derivative operation.
For instance, with the integrand \( \frac{1}{\sqrt{u}} \) in the definite integral \( \int_{1}^{4} \frac{1}{\sqrt{u}} \, du \), we can rewrite it as \( u^{-1/2} \). The antiderivative of \( u^{-1/2} \) is \( 2u^{1/2} \).
To understand this, remember:
For instance, with the integrand \( \frac{1}{\sqrt{u}} \) in the definite integral \( \int_{1}^{4} \frac{1}{\sqrt{u}} \, du \), we can rewrite it as \( u^{-1/2} \). The antiderivative of \( u^{-1/2} \) is \( 2u^{1/2} \).
To understand this, remember:
- Differentiate \( u^{n} \) yields \( n \cdot u^{n-1} \).
- To find an antiderivative, we do the reverse: increase the exponent by 1 and divide by the new exponent.
Other exercises in this chapter
Problem 53
Use integration by substitution and the Fundamental Theorem to evaluate the definite integrals. $$\int_{1}^{4} \frac{e^{\sqrt{x}}}{\sqrt{x}} d x$$
View solution Problem 53
Find the indefinite integrals. $$\int\left(1+\frac{1}{p}\right) d p$$
View solution Problem 54
Find the indefinite integrals. $$\int t^{12} d t$$
View solution Problem 55
Use integration by substitution and the Fundamental Theorem to evaluate the definite integrals. $$\int_{-1}^{e-2} \frac{1}{t+2} d t$$
View solution