in this video we're going to focus on evaluating definite integrals using u substitution so let's start with this example let's say if we want to find the value of the definite integral of 2x x squared plus four raised to the second power from zero to two so first we need to integrate the function using u substitution so we're going to make u equal to x squared plus 4 so that d u is 2x which can get rid of the 2x in the front now du is going to be 2x times dx and we need to divide both sides by 2x to isolate dx so let's replace u or x squared plus 4 with u and let's replace the dx with du over two x so this is going to be u to the second power and d u over two x now as soon as you get rid of all of the x variables and replace with you variables you need to adjust the lower limits and the upper limits using this expression so what is u when x is zero so let's plug in zero zero squared plus four is four so the lower limit is four now the upper limit as an x value is two what is it as a u value so two squared plus four that's four plus four so this is going to be eight so now we can evaluate u squared from 4 to 8. the antiderivative of u squared is going to be u to the third over three evaluated from four to eight so if we plug in eight it's gonna be eight to the third over three and then minus four to the third over three when we plug in four so now what is eight cubed so that's eight times eight times eight which is five hundred and twelve four to the third power is sixty-four and five-twelve minus sixty-four is 448 which is not divisible by 3. so this is the final answer 448 over 3 and so that's how you can evaluate definite integrals using u substitution as soon as you change the variables from x to u make sure to change the value of the lower limit and the upper limit from x-values to their corresponding u-values here's another problem that you could try go ahead and evaluate this particular definite integral using u substitution so we're going to set u equal to this expression because x squared is higher than x now if u is equal to 16 minus x squared what is d u the derivative of sixteen is zero and the derivative of negative x squared is negative two x and now let's solve for dx so let's divide by negative two x so dx is equal to du divided by negative 2x and now at this point let's replace 16 minus x squared with the u variable and let's replace dx with du over negative 2x so this is going to be 4x times the square root of u and then d u divided by negative two x so four x divided by negative two x that's going to be negative two and now we need to change these values so let's plug it into this expression so when x is 0 what is the value of u that's 16 minus 0 squared and so this is going to be 16. now when x is 4 what is u 16 minus 4 squared that's 16 minus 16 that's gonna be zero and the square root of u is u to the one half so now we don't need this anymore right now now what is the antiderivative of u raised to the one half so one half plus one that's three over two and instead of dividing by three over two we're going to multiply by the reciprocal two over three and we still have a negative two in front and let's evaluate this from 16 to zero so negative two times two thirds that's negative four thirds so if we plug in zero it's going to be negative four over three times zero to the three halves and then minus negative four over three times sixteen raised to the three over two so zero raised to anything is just zero and here we have two negative signs so that's going to become positive so we have positive four over three times sixteen raised to the three over two now what is sixteen raised to the one point five or three over two how can we evaluate this rational exponent this is the same as 16 raised to the one half raised to the three when you raise one exponent to another you need to multiply it sixteen to the one half is the same as the square root of sixteen which we know to be four and four to the third is sixty-four so we now have four over three times sixty-four and four times sixty-four four times sixty is two forty because four times six is twenty-four four times 4 is 16 so 240 plus 16 is 256 and therefore this is the final answer for this example it's 256 divided by 3. let's work on one more example evaluate the definite integral from one to two of two x one plus x squared raised to the third power dx now we need to make u equal to 1 plus x squared because d u the derivative of 1 plus x squared is going to be 2x times dx and so we could cancel the 2x on top so now let's solve for dx let's divide both sides by 2x so dx is equal to du over 2x now let's replace one plus x squared with u and let's replace dx with du over 2x so this is going to be 2x divided by u to the third and dx is du over 2x and so we could cancel 2x and don't forget we need to change the limits of integration so what is u when x is one it's going to be one plus one squared which is two now what is the value of u when x is two so it's one plus two squared or one plus four which is five so we need to integrate one over u to the third from two to five so first let's rewrite this expression let's bring the u variable to the top so it's going to be u to the negative three and now the anti-derivative of u to the negative 3 that's going to be negative 3 plus 1 that's negative 2 divided by negative 2 evaluated from 2 to 5. now let's rewrite it so this is negative 1 divided by 2 u squared from two to five and now let's plug in the upper limit so it's negative one over two times five squared and then minus negative 1 over 2 times the lower limit 2 squared 5 squared is 25 and these two negative signs will change into a positive sign 2 squared is 4. now 2 times 25 that's 50. and 2 times 4 is 8 but i'm going to rewrite it as 1 over 8 minus 1 over 50. so we need to get common denominators what number goes into what number is divisible by eight and fifty the only thing i can think of is two hundred eight doesn't go into a hundred but eight goes into two hundred because a goes into forty and forty goes into two hundred so to get 50 to 200 we need to multiply the top and bottom by four two hundred divided by eight is twenty-five so the first fraction we need to multiply the top and bottom by twenty-five so this is gonna be twenty-five over two hundred minus four over two hundred and twenty five minus four is twenty one so the final answer is twenty one divided by two hundred you