How about this integral? Can you evaluate it?michonamona said:Homework Statement
[tex]\int^{\infty}_{0} e^{-y} dy = 1[/tex]
Homework Equations
The Attempt at a Solution
Why is this equality true? I understand that the integral and derivative of e^y is always e^y, but I can't make out why this definite integral is equal to 1. I graphed it using a graphing application and can see why it converges to a finite number, but I can't work it out step by step.
Thanks, I appreciate your help.
M
michonamona said:ok, suppose u=-y, then
[tex]\int^{\infty}_{0} e^{u} dy = e^{u} ] ^{\infty}_{0} = e^{-y} ] ^{\infty}_{0} [/tex]
By the fundamental theorem of calculus
[tex] e^{-y} ]^{\infty}_{0} = 0 - e^0 = -1 [/tex]
Ok, so where in my solution did I go wrong?
A definite integral is a mathematical concept that represents the area under a curve on a graph. It is denoted by the symbol ∫ and has a lower and upper limit which defines the boundaries of the area being calculated.
Exponential functions, which are functions of the form f(x) = ab^x, can be represented as a continuous curve on a graph. The definite integral of an exponential function represents the area under the curve, which has applications in various fields of science and mathematics.
To solve a definite integral of an exponential function, you can use techniques such as integration by parts or substitution. However, for simple exponential functions, you can use the formula ∫ab^x = a(b^x)/(ln b) to calculate the definite integral.
Definite integrals of exponential functions have various applications in fields such as physics, engineering, and economics. They are used to calculate quantities such as growth rates, compound interest, and radioactive decay.
Yes, definite integrals of exponential functions can have negative values. This occurs when the area under the curve is below the x-axis, which represents a negative value. However, when evaluating a definite integral, the negative value is represented as a positive value with a negative sign in front of it.