You use Riemann sums to find the actual area underneath the graph of f(x).
Example: f(x) = x2 + 1 on [0, 2] with 4 equal subintervals
Left sum is an underestimate.
Right sum is an overestimate.
The area under the curve of derivatives of F from A to B is equal to the change in y-values of the function F from A to B, given f is:
● Continuous in interval [a, b]
● F is any function that satisfies F(x) = f(x)
Given y' or f'(x), the anti-derivative can be thought of as the original function, f(x). Integration
is used to find the original function.
● The operation of finding all solutions to this equation is called antidifferentiation or indefinite integration.
● Detonated by an integral sign: ∫
v = ∫ f '(x) dx = f(x) + c
● f '(x) = integrand
● dx = variable of integration
● f(x) = antiderivative
● c = constant of integration
● ∫ = integral
Reminder: ALWAYS add + C when you’re solving for an INDEFINITE integral!
Reminder: Differentiation and integration are inverses!
*K=constant/number
Examples with trig functions
∫ 2sinx dx = 2 ∫ sinx dx = 2(-cosx + c) = -2cosx + c
C is still a constant when multiplied by 2
● HINT: How I memorize antiderivatives by using derivatives of trigonometric functions.
○ EX: d/dx sinx = cosx and for the antiderivative, you just switch the two trigonometric functions and add +C since it’s an indefinite integration.
○ EX: d/dx cscx = -cscxcotx and for the antiderivative, just switch the two trigonometric functions and add +c since it’s an indefinite integration. Also, if the derivative was negative, then the anti-derivative is also negative!
Reminder: Don’t forget to go back and replace u and add +C
● HINT: For ∫ tan u du, memorize like this: ∫ tan u du = ∫ sin u/cos u du because of the trigonometric identities. After that, just do u-substitution with cos u being u.
● If you work it out, it looks like this:
● With e, it’s just the same thing as regular u-substitution but with the additional ‘e’.
Example: ∫7-xdx → - ∫7udu → -1/ln(7) * 7u + C → -7-x/ln7 + C
● u = -x → du = -dx → -du = dx