# Anti-Derivitives, Definite Intervals, and U-Substitutions

How to evaluate the definite integral with trig and u substitution
How to evaluate the definite integral with trig and u substitution

Anti-derivitives, definite intervals, and u-substitutions are all different ways common in calculus to solve complicated problems. Anti-Derivites are self explanatory, you find the reverse of your answer. You have to think backwards, about what you had to had have in order to get the answer provided.

Definite intervals you can use to find the area between curves or the average value. You will use an anti-derivitave to solve for definite intervals.

U-Substitution is another way of solving anti-derivatives. Instead of trying to completely think backwards and figure it out, you would substitute or replace a challenging part of the problem with u and then start to solve from there. You can sub back in u at the end and solve the anti-derivative.

Here are my worksheets #66 and #67 (I was absent for #60):

## Big Ideas

i. What is the difference between an anti-derivative and a definite integral?

In an anti-derivative, you already have the derivative. Your job is to backwards solve the problem to find the original equation. A definite integral is similar and you use an anti-derivative, but the main difference is that you now have a limit on your function.

ii. Why is d/dx lnx=1x, but ∫ 1/x dx=ln|x|+C?

The derivative of lnx is 1x. We learned this a long time ago going over derivatives. The anti-derivative of 1/x is ln|x| + C. This makes sense because we worked backwards to find our original function, lnx. You may notice some differences though, first we add absolute values around x because we know that x must be positive. And second we add +C because their could always be a constant added to the function and we may never see it because the derivative of a constant is 0.

iii. Describe your approach to finding to finding an anti-derivative via u-substitution. Illustrate your process by explaining how you find ∫ cosx/sinx dx?

For u-substitution you can start out by taking out a part of the problem and entirely setting it equal to U, and then substituting. So for this problem I substituted sinx for u, so you get ∫ cosx/u dx. Because the derivative of sinx = cosx we can substitute du for (cosx)(dx), giving us du/u. Because we are still taking the anti-derivate, we get ln|u| +c because that is the anti-derivate of du/u. Remember we add c just in case of the possibility that there was a constant. Now finally we can substitute sinx = u back in replace of you, giving you a final answer of ln|sinx| + c.

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