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### Course: Calculus, all content (2017 edition)>Unit 4

Lesson 3: Indefinite integrals of common functions

# Particular solutions to differential equations: exponential function

Sal finds f(0) given that f'(x)=5eˣ and f(7)=40+5e⁷.

## Want to join the conversation?

• At , how is it allowed to 'take a constant out of the integral sign'?
• Since integral is in a way an infinite sum, 5 is constant with each term (it doesn't change) so you can just factor it out.
• What if you want to integrate an equation that has the variable in its exponent, and is also being multiplied with a constant?
Example:Integration of (0.67*e^0.044x)
• Well, you can take 0.67 out of the integral, then you can use u-substitution for e^0.044x, setting u =
0.044x. Then du = 0.044dx, or dx = du/0.044 = (1/0.044)du. Then the integral becomes 0.67∫(e^u)*(1/0.044)du. You can take 1/0.044 out of the integral since it is a constant. The integral of e^u is e^u. But you need to unsubstitute the u, so the answer is (0.67/0.044)*e^0.044x, or 15.227e^0.044x. Try taking the derivative of this to double check! Hope this made sense.
• I know the d(variable) in the antiderivative is used to let you know with respect to what you are antiderivating, but does it serve any other purpose? Or is there any other reason for it to be there?
• Good question. There is another reason, actually, and it stems from the early idea of an integral.

Remember when we treated the area under a curve as a summation of infinite, thin rectangles? Now, the area of a rectangle is length * width. The length of the rectangle was essentially the distance from y = 0 to the function. So, the length was the value of the function f(x). Now, the width was a small change in the x direction. So, we take it as Δx. We have the length and width. So, area of one rectangle = f(x)Δx.

Now, if the number of rectangles tends to infinity, Δx tends to 0. Whenever something tends to 0, Δ becomes d. So, area of a very, very small rectangle = f(x)dx. Now, as there are infinitely many, we integrate and hence, area under the curve = int(f(x)dx)

So, while dx gives you the variable of integration, it is also the width of a rectangle with infinitesimally small area.
• hi, one of the questions in the next test asks "dy/dx = 3y and y(0)=3. Find y(ln2)".

I'm quite confused about how dy/dx can be expressed in terms of y, rather than x. I'm not sure how to interpret it. Thanks.
(1 vote)
• In cases where dy/dx has a y in it, the y itself would be a function of x. So, think of this as dy/dx = 3y(x). That should make more sense.

You can actually solve for y in this differential equation, differentiate that expression, substitute y in the DE given, and you'll see that they're equal.
• This is a bit of an off topic question, but I am learning integral calculus to comprehensively understand the Basel problem. So what should I target in order to comprehend the famous problem quicker? Thanks
(1 vote)
• Why exercise of exponential is above this video?
(1 vote)
• hi guys after 7 years
(1 vote)
• hey after 10 minutes. lol
(1 vote)
• why not increase the power of e by 1....
in the case of 5e^x anti derivative ?
(1 vote)
• Because e^x is an exponential function, and the power rule is not enough to evaluate its derivative. The power rule is applicable only to some variable in the base of a power and a constant in the exponent of that power. But with e^x the variable is in the exponent, and e (the base) is just constant number!
So we use the chain rule to differentiate e^x. If you're still not sure why, watch the lessons about the power rule and derivative of e^x
(1 vote)
• As Sal does not extract a value from the function f(7)= 40 + 5e^7, I can safely assume e is a variable and not the number e. We have two variables, x and e, and I am assuming a domain of all real numbers. At , Sal says f(0)=5e^0 + 40 = 5(1) +40=45. Sal assumes e to be positive but he no where stated in defining his problem "for all e>0." Likewise, if e is a function of x and x's domain is all real numbers, the range is not inherently limited to positive values for e. My argument is that the answer is undefined as e^0 could be -1 or +1.
• The integral of `e^x` is just that- `e^x`.