I can approximate the area under a curve via left and right endpoints for a given amount of rectangles.
- use summation equations (shown in example with variable definitions in margin) and plug in information to solve
- easy to mess up negatives, forget to add lower limit in the correct place, or use proper equation for left vs right endpoints
- plug in values for variables
- keep the delta x outside of the parentheses (more efficient) and write out the summation of the function with each i value
- add the values together and multiply by delta x to find answer
I can find the area under the curve using Riemann’s Summation.
- use a similar summation equation as with approximations, but use n rectangles and take the limit as n goes to infinity
- important to watch negatives when you break up the summation and remember to distribute any fractions that were taken out of the summation all the way through
- after plugging in values for variables, pull out delta x (just out of summation, not out of limit)
- then distribute any exponents so that the summations can be broken up
- break apart summations anywhere there is addition or subtraction to have multiple summations
- pull anything possible out of each summation until left with either a constant or i (could also be i squared)
- complete the summation, watching to distribute delta x appropriately and maintain any negatives
- multiply fractions, breaking any with addition or subtraction in the numerator apart
- take the limit as n goes to infinity to find answer
I can use U-Substitution to integrate functions.
- substitute u for a piece of the function to make integration easier, plug in given interval to find specific value rather than equation
- vital to remember to multiply any fractions back in to get the correct answer and plug the interval into whatever u equals to have the correct interval once the equation has u as the variable instead of x (if limits aren't given, it is really important to add a constant if you don't want to lose points on a quiz)
- choose a value for u that will make the equation easier to solve, in this case 2x-5
- find du, then multiply the function by any necessary value to make sure du appears in the equation (if the function is multiplied by something, a value is needed outside of the integral to negate this change to the function: multiply the integral by one half if the function must be multiplied by two)
- plug the upper and lower limit into u to find new interval
- plug u and du into the equation
- take the antiderivative of the function
- plug the upper limit into the antiderivative and subtract the antiderivative with the lower limit plugged in to find the answer
- ****if limits are not given, plug u back in so that the original variable is back and don't forget to add a constant at the end of the antiderivative
