Find the area under the graph of f(x)=e−x1+e−xf\left(x\right)=\frac{e^{-x}}{1+e^{-x}} from x=0x=0 to x=4x=4.

ln ⁡ ( 2 1 + e − 4 ) ≈ 0.67 \ln\left(\frac{2}{1+e^{-4}}\right)\thickapprox0.67

Find the area under the graph of f ( x ) = e − x 1 + e − x f\left(x\right)=\frac{e^{-x}}{1+e^{-x}} from x = 0 x=0 to x = 4 x=4 .

Here we're asked to find the area under the graph of the function f of x equals e to the power of negative x over one plus e to the power of negative x from x equals zero to x equals four. Now remember that the area under a function's curve is just equal to the definite integral. So that's exactly what we're gonna be doing here, finding the definite integral from zero to four, those bounds given to us in our problem, of our function e to the negative x over one plus e to the negative x, dx. Now looking at this here, I know that I'm going to make a substitution in order to take this integral. So I want to go ahead and set this denominator equal to u. So if u is equal to one plus e to the power of negative x, du is then going to be equal to negative e to the negative x dx. Now we can see that in our integral here, we have e to the negative x dx. We're just missing that extra negative there. So if I go ahead and multiply my entire integral by negative one and I multiply the inside of my integral by negative one, now I have du along with that u there. So I can rewrite this as negative integral one over u du. Now I'm just gonna ignore my bounds for now until I put this back in terms of x. So let's start by taking this integral here, or let's continue by taking this integral. Now negative integral one over u du, this is going to be negative natural log of the absolute value of u plus c. Now, ultimately, here, we don't have to worry about that plus c, and we can go ahead and plug that u back in here because we know that it's one plus e to the negative x. So the negative natural log of the absolute value of one plus e to the negative x. Having put that back in terms of x, I can now bound this with my original bounds from zero to four. So let's go ahead and plug these bounds in here. This is negative. And I keep that negative out front. Natural log of one plus e to the power of negative four and then minus the natural log of the absolute value of one plus e to the power of zero. Now e to the power of zero is one. So I can rewrite this as negative natural log of one plus e to the power of negative four. Now I don't actually have to worry about my absolute value signs here because that's always going to be a positive value And then minus the natural log of two, I also don't have to worry about absolute value there because two is already positive as well. Now looking at this here, we can go ahead and rewrite some stuff here simplifying this based on log properties. So that's exactly what I'm going to do here. If I distribute this negative in here, this is going to be negative natural log of one plus e to the power of negative four and then plus the natural log of two. Now we can combine those two natural logs into one because if I'm subtracting here that means that I can put it on a fraction or on the bottom of a fraction rather. So the natural log of two over one plus e to the power of negative four. So this is all condensed down into one natural log. So this is an appropriate answer at this point. But if you are actually using a calculator and you want this as a decimal, you can type this right here into your calculator and you should ultimately get here a value of 0.67 for that final answer, the area under our curve. So either of these are correct answers just depending on what's expected of you in your course and whether or not you're using a calculator to get decimal values. Feel free to let us know if you have any questions, and I'll see you in the next one.

Find the area under the graph of f ( x ) = e − x 1 + e − x f\left(x\right)=\frac{e^{-x}}{1+e^{-x}} from x = 0 x=0 to x = 4 x=4 .

ln ⁡ ( 2 1 + e − 4 ) ≈ 0.67 \ln\left(\frac{2}{1+e^{-4}}\right)\thickapprox0.67

ln ⁡ ( 2 e − 4 1 + e − 4 ) ≈ − 3.33 \ln\left(\frac{2e^{-4}}{1+e^{-4}}\right)\thickapprox-3.33

e ln ⁡ 2 ≈ 1.88 e\ln2\thickapprox1.88

ln ⁡ ( 1 1 + e − 4 ) ≈ − 0.018 \ln\left(\frac{1}{1+e^{-4}}\right)\thickapprox-0.018

11. Integrals of Inverse, Exponential, & Logarithmic Functions

Integrals Involving Logarithmic Functions

Calculus

11. Integrals of Inverse, Exponential, & Logarithmic Functions

Integrals Involving Logarithmic Functions

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Multiple Choice

Find the area under the graph of $f (x) = \frac{e^{- x}}{1 + e^{- x}}$ from $x equals 0$ to $x equals 4$ .

$\ln (\frac{2}{1 + e^{- 4}}) \approx 0.67$

$\ln (\frac{2 e^{- 4}}{1 + e^{- 4}}) \approx - 3.33$

$e l n 2 almost equals 1.88$

$\ln (\frac{1}{1 + e^{- 4}}) \approx - 0.018$

Verified step by step guidance

Step 1: Recognize that the problem involves finding the area under the curve of the function f(x) = e^(-x) / (1 + e^(-x)) from x = 0 to x = 4. This requires evaluating the definite integral of f(x) over the interval [0, 4].

Step 2: Write the integral to be solved: ∫[0 to 4] (e^(-x) / (1 + e^(-x))) dx. This integral represents the area under the curve of f(x) from x = 0 to x = 4.

Step 3: Simplify the function f(x) = e^(-x) / (1 + e^(-x)). Notice that this function can be rewritten as f(x) = 1 / (e^(x) + 1) by multiplying numerator and denominator by e^(x). This simplification may make the integration process easier.

Step 4: Recognize that the integral of 1 / (e^(x) + 1) is a standard integral. The antiderivative of this function is ln(1 + e^(-x)), which can be derived using substitution techniques. Specifically, let u = 1 + e^(-x), then du = -e^(-x) dx.

Step 5: Evaluate the definite integral by substituting the limits of integration (x = 0 and x = 4) into the antiderivative ln(1 + e^(-x)). Compute ln(1 + e^(-4)) - ln(1 + e^(0)). Simplify the expression to find the final result.

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