in this video we're going to focus on evaluating the definite integral but first what is the difference between an indefinite integral and the definite integral so on the left i have an indefinite integral and on the right i have a definite integral the difference is a definite integral containing the lower limit a and the upper limit b those are known as the limits of integration the indefinite integral doesn't happen now once you integrate the indefinite integral it's going to give you a function in terms of x based on this example so this will give you the anti-derivative of lowercase f which is capital f plus c when you integrate the definite integral you're going to get a number maybe like 25 or something so it gives you a specific value f of x is known as the integrand dx is the variable of integration and this is the integral sign which represents a limit of sums now you need to be familiar with the fundamental theorem of calculus part two and it's associated with definite integrals basically the definite integral of f from a to b is equal to f of b minus f of a where capital f is the anti-derivative of lowercase f now this symbol if you see it is the same as f of b minus f of a so let's go ahead and apply ftc part two the second part of the fundamental theorem of calculus to evaluate definite integrals so let's start with a simple problem let's evaluate eight dx from two to five so first we need to find the anti-derivative of eight the anti-derivative of eight is eight x because the derivative of eight x is eight and we're going to evaluate it from two to five so first let's replace x with five this represents f of b or basically f of five and then we're going to replace x with two so this is the f of a part now 8 times 5 is 40 and 8 times 2 is 16 40 minus 16 is 24. so the final answer of any definite integral is a number and that's it for the first example now let's try another problem go ahead and evaluate this particular definite integral feel free to pause the video if you want to so first let's find the anti-derivative of five x minus four so we need to use the power rule the anti-derivative of x raised to the n is x raised to the n plus one divided by n plus one now anytime you have an indefinite integral you're going to get some constant c when you have a definite integral you don't need the constant of integration it's going to cancel so the antiderivative of 5x is going to be 5 x squared divided by two and for negative four it's going to be negative four x now let's see what would happen if we add c which we really don't need to but let's just see what's going to happen so first we need to calculate f of b so we need to plug in 4 into this entire expression so it's going to be 5 times 4 squared divided by 2 minus 4x or 4 times 4 plus c and then minus now f of a so we got to plug in 1 this time so it's going to be 5 times 1 squared over 2 minus 4 times 1 plus c now you can see why c is about to cancel it doesn't have an x variable attached to it 4 squared is 16 16 divided by 2 is 8. 8 times five is forty four times four is sixteen and then on the right side we have one squared times five over two so that's five over two and don't forget to distribute the negative sign to everything on the right and then we have negative 4 times 1 that's negative 4 times the negative sign so that's going to be positive 4 and then we're going to have negative c so c and negative c will cancel that's why it's not needed 40 minus 16 that's going to be 24 and 24 plus 4 is 28 now 5 over two i can break that into four over two minus a half because negative four minus one is negative five and four divided by two is two so i have 28 minus two which is 26 and so i have 26 minus a half so as a mixed number that's going to be 25 and one half or 25.5 as an improper fraction is going to be 25 times 2 which is 50 plus 1 and keep the denominator the same so it's going to be 51 over 2. so this is the final answer for this problem let's work on this one go ahead and integrate this function from negative three to four so the first thing we need to do is rewrite the expression so let's take the x variable and move it to the top before we integrate it so we have 8 x to the negative 3 dx so now let's find the anti-derivative of this expression using a power rule so negative 3 plus 1 that's going to be negative 2 and then we need to divide by it and then let's evaluate this from negative 3 to 4. but before we do that let's simplify and rewrite this expression so 8 divided by negative 2 that's going to be negative 4 and then we could take the x variable and move it back to the bottom so it's going to change from x raised to the negative 2 to x raised to the positive 2. and so now we can plug in four and three so let's start with the top one the upper limit so it's going to be negative four over four squared so that's f of b and then minus f of a that's going to be negative four over negative three squared so negative four over four squared we can cancel a four so we're left with negative one over four on the right side these two negative signs will cancel and so that's gonna be positive four on top negative three squared is positive nine so now we need to add these two fractions so what we need to do is get common denominators so a common denominator between 4 and 9 is going to be 36. so we're going to multiply the second fraction by 4 over 4 and the first one by 9 over 9. so this is going to be negative 9 over thirty six plus sixteen over thirty six sixteen minus nine is seven so the final answer is seven divided by thirty six and that's the answer try this example evaluate the definite integral from 1 to e of 5 divided by x dx so first let's rewrite it as 5 and times 1 over x i'm going to move the constant to the front now what is the anti-derivative of 1 over x dx if you recall that's ln x so what we have is 5 times ln x evaluated from 1 to e so f b is going to be 5 ln e and then minus f of a which is 5 ln 1. now i need to know that the natural log of e is one and the natural log of one is zero so the final answer is simply five and that's it for this problem it's not too difficult let's try one more problem let's find the value of this definite integral go ahead and try this so first we need to rewrite the expression the square root of x is the same as x to the one-half next we need to move x to the top so this is going to be x raised to the negative one-half and now we can use the power rule so negative one-half plus one that's going to be positive one-half now and then we need to divide by one-half but if you multiply the top and bottom by two two times a half is one so this whole expression becomes two x to the one half evaluated from four to nine and then we can convert it back into the square root symbol so f of b is going to be 2 times the square root of 9 minus f of a which is 2 times the square root of 4. the square root of nine is three and the square root of four is two two times three is six two times two is four six minus four is two and that's the final answer you