Problem 106
Question
In the following exercises, use the comparison theorem. Show that \(\int_{0}^{1} \sqrt{1+x^{3}} d x \leq \int_{0}^{1} \sqrt{1+x^{2}} d x\)
Step-by-Step Solution
Verified Answer
The integral of \( \sqrt{1+x^3} \) is less than or equal to the integral of \( \sqrt{1+x^2} \) over \([0, 1]\).
1Step 1: Understand the Comparison Theorem
The Comparison Theorem is used to compare the values of two integrals when one function is always greater than or equal to another over a given interval. If \( f(x) \leq g(x) \) for all \( x \) in the interval, then \( \int f(x) \,dx \leq \int g(x) \,dx \).
2Step 2: Compare the Functions
We need to compare \( \sqrt{1 + x^3} \) and \( \sqrt{1 + x^2} \). We observe that for every \( x \) in the interval \([0, 1]\), \( x^3 \leq x^2 \). Thus, \( 1 + x^3 \leq 1 + x^2 \). Taking the square root of both sides, we have \( \sqrt{1 + x^3} \leq \sqrt{1 + x^2} \).
3Step 3: Apply the Comparison
Since \( \sqrt{1 + x^3} \leq \sqrt{1 + x^2} \) for all \( x \) in \([0, 1]\), we can use the Comparison Theorem to conclude that \( \int_{0}^{1} \sqrt{1+x^{3}} \,dx \leq \int_{0}^{1} \sqrt{1+x^{2}} \,dx \).
Key Concepts
Definite IntegralsFunction ComparisonIntegral Inequalities
Definite Integrals
Definite integrals allow us to calculate the area under a curve over a specific interval. This is vital in mathematics because it helps in finding out the total accumulation or net area from a function over an interval. When you see a notation like \( \int_{a}^{b} f(x) \, dx \), it is asking you to find the area under the curve of \( f(x) \) from \( x = a \) to \( x = b \). This is done by summing up infinitely small areas, which can be thought of as narrow rectangles, under the curve between these bounds.
When performing these calculations, having a good grasp of how the behavior of the function affects the integral is crucial. This involves understanding where the function increases or decreases, and how these changes affect the total area accumulated. In the context of the original exercise, we are looking at definite integrals of two different functions over the same interval from 0 to 1, and comparing their values.
When performing these calculations, having a good grasp of how the behavior of the function affects the integral is crucial. This involves understanding where the function increases or decreases, and how these changes affect the total area accumulated. In the context of the original exercise, we are looking at definite integrals of two different functions over the same interval from 0 to 1, and comparing their values.
Function Comparison
Function comparison forms the backbone of the Comparison Theorem. This theorem is a powerful tool because it gives a way to estimate the integral of a function by using another function that is easier to integrate. The idea is to look at the values of two functions over the same interval and establish which one dominates or is greater.
In our exercise, we need to compare the functions \( \sqrt{1+x^3} \) and \( \sqrt{1+x^2} \). By examining the expressions inside the square root, we notice that for all \( x \) in the interval \([0, 1]\), \( x^3 \) is less than or equal to \( x^2 \). Thus, the function \( \sqrt{1+x^3} \) turns out to be smaller than or equal to \( \sqrt{1+x^2} \), confirming the comparison.
In our exercise, we need to compare the functions \( \sqrt{1+x^3} \) and \( \sqrt{1+x^2} \). By examining the expressions inside the square root, we notice that for all \( x \) in the interval \([0, 1]\), \( x^3 \) is less than or equal to \( x^2 \). Thus, the function \( \sqrt{1+x^3} \) turns out to be smaller than or equal to \( \sqrt{1+x^2} \), confirming the comparison.
Integral Inequalities
Integral inequalities are directly tied to the Comparison Theorem and are used to establish a relationship between two integrals. If one function is always less than or equal to another over an interval, then by integrating, you can establish an inequality for their definite integrals. This is extremely useful because it simplifies the process of finding the bounds for an integral.
In the exercise at hand, applying integral inequalities means showing that since \( \sqrt{1+x^3} \leq \sqrt{1+x^2} \) for all \( x \) in \([0, 1]\), then the integral of \( \sqrt{1+x^3} \) from 0 to 1 is less than or equal to the integral of \( \sqrt{1+x^2} \) over the same interval. This inequality simplifies understanding how the areas under these curves compare to each other.
In the exercise at hand, applying integral inequalities means showing that since \( \sqrt{1+x^3} \leq \sqrt{1+x^2} \) for all \( x \) in \([0, 1]\), then the integral of \( \sqrt{1+x^3} \) from 0 to 1 is less than or equal to the integral of \( \sqrt{1+x^2} \) over the same interval. This inequality simplifies understanding how the areas under these curves compare to each other.
Other exercises in this chapter
Problem 104
In the following exercises, use the comparison theorem. Show that \(\int_{0}^{3}\left(x^{2}-6 x+9\right) d x \geq 0\)
View solution Problem 105
In the following exercises, use the comparison theorem. Show that \(\int_{-2}^{3}(x-3)(x+2) d x \leq 0\)
View solution Problem 107
In the following exercises, use the comparison theorem. Show that \(\int_{1}^{2} \sqrt{1+x} d x \leq \int_{1}^{2} \sqrt{1+x^{2}} d x\)
View solution Problem 108
In the following exercises, use the comparison theorem. Show that \(\int_{0}^{\pi / 2} \sin t d t \geq \frac{\pi}{4} .\) (Hint: sin \(t \geq \frac{2 t}{\pi}\) o
View solution