Find the area of f ( x ) = 3 + e 2 x f\left(x\right)=3+e^{2x} from x = 0 x=0 to x = 2 x=2 .

this problem, we're asked to find the area of the function f of x equals three plus e to the power of two x from x equals zero to x equals two. Now remember that to find the area under the curve of a function, we have to take an integral. So here we're going to be integrating from zero to two. Those bounds were given to us here. And then we're integrating our function three plus e to the power of two x d x. So taking this integral here, that antiderivative of three is just going to be three x. Then looking at the second term here, e to the power of two x. Just to fully break this down and taking this integral, we would need to be doing a substitution where u is equal to two x. Now if u is equal to two x, that makes du equal to two d x. And since we're missing that two here, that means we ultimately are going to have to multiply by two as well as one half in order to make this integral a one half, the integral of e to the u du. Then this will ultimately give us a one half e to the u. And we know that u is just two x, so we can plug that back in. So this is my second term here. I'm gonna go ahead and bring that up. So I have my three x then having integrated that e to the two x to get one half E to the two X and then bounded from zero to two. So now let's go ahead and just plug in our bounds. So plugging in our bounds here, this is going to be three times two plus one half E to the power of two times two. That's my upper bound plugged in. Then subtracting off our lower bound. This is three times zero plus one half e to the power of two times zero. Now three times zero is zero and e to the power of zero is just going to be one. So we can do some simplification here. Three times two is six. So this is six plus e to the power of four over two minus one half is all that's left over here. Then we can go ahead and give this six a common denominator of two by multiplying it by two over two. So that gives me here 12 plus e to the power of four minus one and all of that is now over two. So now we can combine our like terms that 12 and that negative one gives us 11. So this is 11 plus e to the power of four over two and this is my final answer if I can't use a calculator. Now if you can use a calculator, you would wanna plug this in in order to get your final answer in decimal form. And when you do that, you should end up getting here thirty two point eight as your final answer. Feel free to let us know if you have any questions in the comments, and I'll see you in the next one.

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11. Integrals of Inverse, Exponential, & Logarithmic Functions

Integrals of Exponential Functions

Calculus

11. Integrals of Inverse, Exponential, & Logarithmic Functions

Integrals of Exponential Functions

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Multiple Choice

Find the area of $f (x) = 3 + e^{2 x}$ from $x equals 0$ to $x equals 2$ .

-20.80

32.80

-21.80

-21.30

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Verified step by step guidance

Set up the definite integral to find the area under the curve. The area is given by the integral of f(x) = 3 + e^{2x} from x = 0 to x = 2. This can be written as: ∫[0,2] (3 + e^{2x}) dx.

Split the integral into two parts for simplicity: ∫[0,2] 3 dx + ∫[0,2] e^{2x} dx.

Evaluate the first integral, ∫[0,2] 3 dx. Since 3 is a constant, the integral becomes 3x, and you will evaluate it from x = 0 to x = 2.

For the second integral, ∫[0,2] e^{2x} dx, use the substitution method. Let u = 2x, so du = 2 dx, or dx = du/2. Rewrite the integral in terms of u and solve. After solving, substitute back to express the result in terms of x and evaluate it from x = 0 to x = 2.

Add the results of the two integrals together to find the total area under the curve from x = 0 to x = 2.