TheForce said:Find the maximum value of the function on the interval and multiply that by the interval length, tat is the maximum possible value for the integral (the upper bound). Try that and let us know how it goes.
SqueeSpleen said:I don't know how exactly they want you to solve this, but you can find functions that are greater than your's in the integration interval.
For example:
√(x+e^-x) ≤ √3 in (0,2) and √(x+e^-x) ≤ √4 in (2,3).
So ∫(0,3) √(x+e^-x)dx ≤ 2√3+2
This bounding is awful but I think it's useful as example.
LCKurtz said:I think you will have good luck by overestimating the e-x and evaluating the resulting integral.
You have found a rather crude bound by drawing a horizontal line at ##y = \sqrt{3 + e^{-3}}##. But the function value is much smaller at ##x = 0##: ##\sqrt{0 + e^0} = 1##. Try taking advantage of that fact, for example, by drawing a line from the point ##(x=0, y=1)## to ##(x = 3, y=\sqrt{3 + e^{-3}})##. Does your function stay below that line? If so, that will give you a smaller bound.dillon131222 said:need to show that the integral is ≤ 14/3, so don't see how showing that its ≤ 2√3+2 which = 5.46 helps(?) or am i misunderstanding?
dillon131222 said:need to show that the integral is ≤ 14/3...
not sure what you mean by overestimating e-x
jbunniii said:You have found a rather crude bound by drawing a horizontal line at ##y = \sqrt{3 + e^{-3}}##. But the function value is much smaller at ##x = 0##: ##\sqrt{0 + e^0} = 1##. Try taking advantage of that fact, for example, by drawing a line from the point ##(x=0, y=1)## to ##(x = 3, y=\sqrt{3 + e^{-3}})##. Does your function stay below that line? If so, that will give you a smaller bound.
LCKurtz said:What is the largest value ##e^{-x}## attains on [0,3]? Overestimate the integral by replacing ##e^{-x}## with that largest value. It will give you a simpler integral you can work.
The purpose of proving this inequality is to demonstrate a fundamental concept in Calculus II, which is the comparison between the area under a curve and a specific value. This proof also helps to solidify understanding of integration and its applications.
To prove this inequality, one can use various methods such as the comparison test, the limit comparison test, or the integral test. These methods involve comparing the given integral to a known integral with a known value, and using properties of integrals to show that the given integral is less than or equal to the known value.
This inequality can be proven for any continuous function on the given interval, as long as the function is non-negative. This includes polynomial functions, exponential functions, trigonometric functions, and more.
No, it is not necessary to have prior knowledge of integration techniques to prove this inequality. However, a basic understanding of integration and its properties is helpful in understanding the proof and being able to follow the steps.
Yes, there are several real-world applications of proving this inequality. For example, it can be used to approximate the value of a definite integral without having to evaluate it directly. It can also be used in economics and physics to calculate quantities such as work, volume, and area. Additionally, it is an important concept in optimization problems where the goal is to find the maximum or minimum value of a function.