Calculus - Evaluating a definite integral

In this video I cover the basic idea behind evaluating a definite integral.
This is really using the fundamental theorem of calculus part 2. Remember to take an entire step just to find the anti-derivative. This will really help with the rest of your calculations later one.
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This video is related to many other topics. Check them out:
The basics of indefinite integrals: • Calculus - The basics ...
Find the derivative of inverse trigonometric functions: • Calculus - Find the de...
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[@laonethebe9203]

why don't we add a constant when finding the anti-derivative for definite integral?

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I'm pretty sure if you add C it will cancel out because it will be the same in F(b) and F(a)

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[@justabunga1]

I see in the second example. You could have done using properties of integral for even and odd functions of the limits of integration with the same endpoint but with different signs. For the integral of 1/(1+x^2) dx from -1 to 1, you can rewrite this as 2*integral of 1/(1+x^2) dx from 0 to 1 and then go from there. For even functions (i.e. f(-x)=f(x)), it is rewritten as 2*integral of a function. For odd functions (i.e. f(-x)=-f(x)), the answer is 0. There are some functions that can't be integrated in terms of a function (e.g. e^(-x^2), xtan(x), 2^cos(x), etc.). Whenever you see a function with the same endpoints for the limits of integrations such as an odd function even though a function is not integrable, the answer is always 0. A good example of this would be the integral of tan(x)/(1+x^4) from -1 to 1 dx is 0. For even functions, the answer is non-elementary.

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When you're writing down an integral it's better to put the quantity in parentheses and leave the differential dx outside. This way it becomes evident that dx goes to both terms of the function to integrate.

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