Problem 14
Question
use the Second Fundamental Theorem of Calculus to evaluate each definite integral. $$ \int_{0}^{1}\left(x^{4 / 3}-2 x^{1 / 3}\right) d x $$
Step-by-Step Solution
Verified Answer
The value of the integral is \(-\frac{15}{14}\).
1Step 1: Understand the Integral
We need to evaluate the integral \( \int_{0}^{1}(x^{4/3} - 2x^{1/3}) \, dx \). This involves finding the antiderivative of the function \( f(x) = x^{4/3} - 2x^{1/3} \) and then applying the limits from 0 to 1.
2Step 2: Find the Antiderivative
The antiderivative of \( x^{n} \) is \( \frac{x^{n+1}}{n+1} \). Therefore, the antiderivative of \( x^{4/3} \) is \( \frac{x^{4/3+1}}{4/3+1} = \frac{x^{7/3}}{7/3} = \frac{3}{7}x^{7/3} \). For \( -2x^{1/3} \), the antiderivative is \( -2\frac{x^{1/3+1}}{1/3+1} = -2\frac{x^{4/3}}{4/3} = -\frac{6}{4}x^{4/3} = -\frac{3}{2}x^{4/3} \). Thus, the antiderivative of \( f(x) \) is \( F(x) = \frac{3}{7}x^{7/3} - \frac{3}{2}x^{4/3} \).
3Step 3: Apply the Limits of Integration
Using the Second Fundamental Theorem of Calculus, evaluate the definite integral by calculating \( F(1) - F(0) \). Substitute into the antiderivative: \( F(1) = \frac{3}{7}(1)^{7/3} - \frac{3}{2}(1)^{4/3} = \frac{3}{7} - \frac{3}{2} \). Similarly, compute \( F(0) = \frac{3}{7}(0)^{7/3} - \frac{3}{2}(0)^{4/3} = 0 \). Hence, \( F(1) - F(0) = \left( \frac{3}{7} - \frac{3}{2} \right) - 0 \).
4Step 4: Simplify the Result
Simplify the expression \( \frac{3}{7} - \frac{3}{2} \). To do this, find a common denominator for the fractions. The common denominator of 7 and 2 is 14. Rewrite each fraction: \( \frac{3}{7} = \frac{6}{14} \) and \( \frac{3}{2} = \frac{21}{14} \). Subtract: \( \frac{6}{14} - \frac{21}{14} = -\frac{15}{14} \). Thus, the value of the integral is \(-\frac{15}{14}\).
Key Concepts
Definite IntegralAntiderivativeLimits of IntegrationSimplification of Fractions
Definite Integral
A definite integral represents the area under the curve of a given function, from one point to another. When we talk about the definite integral of a function from \(a\) to \(b\), we denote it as \[ \int_{a}^{b} f(x) \, dx \].
The definite integral essentially measures the net accumulation of the value of the function from the lower limit \(a\) to the upper limit \(b\). This is often visualized as the signed area between the function and the \(x\)-axis.
In the context of our problem, the function \(f(x) = x^{4/3} - 2x^{1/3}\) is integrated from 0 to 1. Using the Second Fundamental Theorem of Calculus, this tells us that we find the exact total change of the function’s antiderivative over this interval.
The definite integral essentially measures the net accumulation of the value of the function from the lower limit \(a\) to the upper limit \(b\). This is often visualized as the signed area between the function and the \(x\)-axis.
In the context of our problem, the function \(f(x) = x^{4/3} - 2x^{1/3}\) is integrated from 0 to 1. Using the Second Fundamental Theorem of Calculus, this tells us that we find the exact total change of the function’s antiderivative over this interval.
Antiderivative
To integrate a function, it's essential to find its antiderivative, also known as the indefinite integral. The antiderivative can be thought of as the inverse operation of taking a derivative. If \(F(x)\) is an antiderivative of \(f(x)\), then by definition, \(F'(x) = f(x)\).
For our exercise, the antiderivative of \(x^{4/3}\) is calculated by increasing the exponent by 1, giving us \(x^{7/3}\), and then dividing by this new exponent. Thus, the antiderivative is \(\frac{3}{7}x^{7/3}\). Similarly, for \(-2x^{1/3}\), the process results in \(-\frac{3}{2}x^{4/3}\).
These calculations help in determining the expression \(F(x)\) for the integral \[ F(x) = \frac{3}{7}x^{7/3} - \frac{3}{2}x^{4/3} \]. By integrating over a defined interval, we can find the area under the curve.
For our exercise, the antiderivative of \(x^{4/3}\) is calculated by increasing the exponent by 1, giving us \(x^{7/3}\), and then dividing by this new exponent. Thus, the antiderivative is \(\frac{3}{7}x^{7/3}\). Similarly, for \(-2x^{1/3}\), the process results in \(-\frac{3}{2}x^{4/3}\).
These calculations help in determining the expression \(F(x)\) for the integral \[ F(x) = \frac{3}{7}x^{7/3} - \frac{3}{2}x^{4/3} \]. By integrating over a defined interval, we can find the area under the curve.
Limits of Integration
The limits of integration are crucial for evaluating a definite integral, as they specify the interval over which you are calculating the area. Given the integral \[ \int_{0}^{1}(x^{4/3} - 2x^{1/3}) \, dx \], the limits are from \(0\) to \(1\).
To solve it, you substitute the upper limit into the antiderivative function \(F(x)\), and then do the same for the lower limit, subtracting the latter from the former.
In this example, we find \(F(1)\) by substituting \(x = 1\) into our antiderivative, yielding \(\frac{3}{7} - \frac{3}{2}\). For \(F(0)\), substituting \(x = 0\) results in 0. Hence, the definite integral evaluates as \[ F(1) - F(0) = \left(\frac{3}{7} - \frac{3}{2}\right) - 0 \].
To solve it, you substitute the upper limit into the antiderivative function \(F(x)\), and then do the same for the lower limit, subtracting the latter from the former.
In this example, we find \(F(1)\) by substituting \(x = 1\) into our antiderivative, yielding \(\frac{3}{7} - \frac{3}{2}\). For \(F(0)\), substituting \(x = 0\) results in 0. Hence, the definite integral evaluates as \[ F(1) - F(0) = \left(\frac{3}{7} - \frac{3}{2}\right) - 0 \].
Simplification of Fractions
Simplifying fractions often involves finding a common denominator to combine them into one fraction. It's an essential step for calculating differences or sums of fractions accurately.
In our scenario, we need to simplify \(\frac{3}{7} - \frac{3}{2}\). We find the least common denominator (LCD), which for 7 and 2 is 14. Convert each fraction:
Thus, the value of the definite integral is \(-\frac{15}{14}\), representing the area under the curve of \(f(x)\) from \(0\) to \(1\).
In our scenario, we need to simplify \(\frac{3}{7} - \frac{3}{2}\). We find the least common denominator (LCD), which for 7 and 2 is 14. Convert each fraction:
- \(\frac{3}{7} = \frac{6}{14}\)
- \(\frac{3}{2} = \frac{21}{14}\)
Thus, the value of the definite integral is \(-\frac{15}{14}\), representing the area under the curve of \(f(x)\) from \(0\) to \(1\).
Other exercises in this chapter
Problem 14
Suppose that \(\int_{0}^{1} f(x) d x=2, \int_{1}^{2} f(x) d x=3, \int_{0}^{1} g(x) d x=-1\) and \(\int_{0}^{2} g(x) d x=4 .\) Use properties of definite integra
View solution Problem 14
Find the average value of the function on the given interval. $$ G(v)=\frac{\sin v \cos v}{\sqrt{1+\cos ^{2} v}} ; \quad[0, \pi / 2] $$
View solution Problem 15
Suppose that \(\sum_{i=1}^{10} a_{i}=40\) and \(\sum_{i=1}^{10} b_{i}=50 .\) Calculate each of the following. $$ \sum_{i=1}^{10}\left(a_{i}+b_{i}\right) $$
View solution Problem 15
Suppose that \(\int_{0}^{1} f(x) d x=2, \int_{1}^{2} f(x) d x=3, \int_{0}^{1} g(x) d x=-1\) and \(\int_{0}^{2} g(x) d x=4 .\) Use properties of definite integra
View solution