This is an example of how to use the fundamental theorem of calculus. The question is, evaluate the integral from negative two to three of x squared plus four x plus five dx. So we use the fundamental theorem of calculus when we evaluate an integral from point A to B. So this bottom value here is going to be our A value and this top value is our B value. This will always be the case, your bottom value will be your A, and your top value will be your B. The actual formula for the fundamental theorem of calculus is the integral from A to B of f of x dx is equal to the anti-derivative evaluated at point B minus the anti-derivative evaluated at point A. So we can start to take our anti-derivative. So just two rules to remind you that you have to use in order to take this anti-derivative is the integral of x to the n dx which is equals to x to the n plus one divided by n plus one, and this works for all integers except for negative one, and then we need to know how to take the integral of a constant and the integral of a constant is just the constant multiplied by x. So for the first term we have x squared so it’s going to be x cubed over three so its going to be one third multiplied by x cubed and then we have four x, and integrated is four x squared over two, so the four divided by two is just two, so I have plus two x squared and then for the last term we have a constant so we just multiply it by x so we have five x and you’ll notice here I’m using square brackets. I use square brackets because I still have to write in my bounds and its proper notation to write square brackets and write your bounds on either side at the end. Another notation you might see is someone write their anti-derivative in round brackets and use a straight line at the end to

represent their bounds. You’ll also notice here that I don’t add the integration constant. When evaluating an integral with bounds, you don’t have to write the integration constant. So now we can start to plug in our values, so just to remind you, this is our B value and this is our A value. So in the formula it says to plug in our B value first, so if we plug in three everywhere we see x, we are going to get one third three cubed plus two times three squared plus five times three and now we’re going to subtract the anti-derivative with our A value plugged in and negative two is our A value. So if we plug in negative two we’re going to get one third negative two cubed plus two times negative two squared plus five times negative two. So now if we simplify this, we’re going to get three cubed which is three times three which is nine and times three again is

twenty seven and twenty seven divided by three is just nine and then three square is nine and multiplied by two is eighteen, and five times three is fifteen, so plus fifteen, and then we’re going to subtract this bracket simplified so negative two cubed is negative eight and over three

is negative eight over three. Negative two squared is four multiplied by two is plus eight and five times negative two is negative ten. So now to simplify these brackets, nine plus eighteen plus fifteen is forty two, and negative eight over three plus eight minus ten is actually negative fourteen over three. But since there’s a negative out front and this is negative as well, it’s going to turn into a plus fourteen over three. I got fourteen over three because you have negative eight over three and then to make eight and ten over the same denominator we multiplied eight times three so we’re going to get minus eight over three plus twenty four over three and then we multiply ten by three so minus thirty over three and when you add those tops you’re going to get negative fourteen and it’s all over three. So now to add forty two to fourteen over three into a fraction, we multiply forty two by three which is a hundred and twenty six over three plus fourteen over three and a hundred and twenty six plus fourteen is one hundred and forty over three. Now I usually just leave it as a fraction but if your professor asks you to put your answers into decimals. This answer is approximately forty six point seven and that’s rounded to one decimal point. So now you can have your therefore statement, so you have therefore the integral from negative two to three of x squared plus four x plus five dx is equal to a hundred and forty over three, or approximately forty six point seven, and that is how you use the fundamental

theorem of calculus.

Thank you very Much for this problem i learnet a lot from it, i know how to solve my own pMath roblem