Find the area under the graph of f(x) = e-ln(x) on the interval [1, 2].
no
You do realize that
f(x) = e^(-lnx) , can be changed to
f(x) = x^-1 = 1/x
so area = ∫(1/x) dx from 1 to 2
= [lnx] from 1 to 2
= ln2 - ln1
= ln2 or appr .6931...
proof:
http://www.wolframalpha.com/input/?i=%E2%88%AB(e%5E(-lnx))+dx+from+1+to+2
shut up Steve
Well, to find the area under the graph, we need to integrate the function. But before we do that, let me tell you a joke: Why don't scientists trust atoms? Because they make up everything!
Now, let's calculate the integral of f(x) = e^(-ln(x)). First, we can simplify the expression using the rules of logarithms. Since e^ln(x) is just x, f(x) simplifies to just 1/x.
Now, the integral of 1/x is ln(x), so let's integrate ln(x) on the interval [1, 2]. But wait, before we do that, let me tell you another joke: Why don't skeletons fight each other? They don't have the guts!
Anyway, back to integrating ln(x). The integral of ln(x) is xln(x) - x + C, where C is the constant of integration. Evaluating this expression on the interval [1, 2], we get:
[2ln(2) - 2] - [ln(1) - 1]
Since ln(1) is 0, the expression simplifies to:
2ln(2) - 2 - (-1)
And that simplifies further to:
2ln(2) - 1
So, the area under the graph of f(x) = e-ln(x) on the interval [1, 2] is 2ln(2) - 1. And there you have it, along with a couple of jokes to keep things light!
To find the area under the graph of f(x) = e-ln(x) on the interval [1, 2], we can use definite integration. The definite integral of a function represents the signed area between the graph of the function and the x-axis.
First, let's find the antiderivative of f(x) = e-ln(x). The antiderivative of e-ln(x) can be found by applying the rule of logarithms. Recall that ln(x) is the natural logarithm of x.
Using the rule of logarithms, we rewrite e-ln(x) as e * e-ln(x) = e * (1/x) = e/x.
The antiderivative of e/x with respect to x is ln|x| + C, where C is the constant of integration.
Next, we can evaluate the definite integral of f(x) = e-ln(x) on the interval [1, 2] using the antiderivative. The definite integral is denoted by ∫ from a to b, where a and b are the endpoints of the interval.
∫[1,2] e-ln(x) dx = [ln|x|] from 1 to 2
Plugging in the values, we have:
[ln|2|] - [ln|1|]
Since ln(1) = 0, the expression simplifies to:
ln(2) - 0 = ln(2)
Therefore, the area under the graph of f(x) = e-ln(x) on the interval [1, 2] is ln(2).
huh? e^(-lnx) = 1/e^(lnx) = 1/x
You can probably handle that, right?