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4122 Get help what does 1 = ap(a)y Video the of 2. following Integral equal?Assessmenth(v)dv and h(v)dv = 8 Due in 9 hours Minui...

Question

4122 Get help what does 1 = ap(a)y Video the of 2. following Integral equal?Assessmenth(v)dv and h(v)dv = 8 Due in 9 hours Minui

41 22 Get help what does 1 = ap(a)y Video the of 2. following Integral equal? Assessment h(v)dv and h(v)dv = 8 Due in 9 hours Minui



Answers

$21-44$ Evaluate the integral.
$$\int_{-1}^{2}(x-2|x|) d x$$

To integrate this, we first want to find the partial fraction decomposition of one over x squared minus nine. Now one over x squared minus nine. This can be written As one over X -3 times x plus three. And so since the denominator has distinct linear factors, then the partial fractions will Have the nominators X -3 and X plus three. And since they're both linear then the enumerators will be constant. Let's say A and B. Now to find A and B want to multiply this by the L c D X -3 times x plus three. And so we have one equal to eight times x plus three Plus B Times X -3. And in here, yes X minus three equals zero. This gives us X equal to three. Then we have one equal to eight times three plus three plus B times zero. This gives us A equal to 1/6. And if let's say X plus three equals zero, that gives us x equal to negative three. Then we have one equal to eight times zero plus B Times -3 -3. And this gives us be equal to -1/6. So we have 1/ X Grade -9. This is equal to 1/6 over X -3 plus negative 1/6 over X plus three which is just equal to 1/6 times one over X -3 one over X plus three. Now integrating both sides with respect to X we have the integral of this D X. This is equal to the integral of 1/6 times one of X -3 minus one over X plus three D X. And so from here we have 1/6 times the integral of one over x minus three. D x minus the integral of one over X plus three G x. Which is equal to 1/6 times L. An absolute value of x minus three minus Ln absolute value of X plus three and then plus C. And by fundamental theorem of calculus, the definite integral from 0 to 1 of one over X squared minus nine. This is equal to 1/6 of L. And absolute value of x minus three minus Ln of X plus three, evaluated from 0 to 1. This gives us one over six times when X is one, that's Ln off The absolute value of 1 -3, That'll be two Ln absolute value of one plus three. That's four -10. We will have LNF three -6 of three. And we will get and here we have zero. So you will get 1/6 times. LNF to minus Elena four is just to LNF two. So this will be 1/6 times negative. L enough to or that's the same as Negative well enough to over six.

No so x minus 12 power three Gis equals two X power of three minus three X choir plus three X minus one hour X choir. This one equals 2 x minus three plus three over. X minus y know where acts choir mm So the integration of the swan just equals to one hour two X Minour integration of three plus Integration of three h X minus integration of our ex choir. So the first one just equals two ex choir over to Those one equals 2 three x. Yeah so three times long next and for the last one we have minus X my next one. Mhm wow we are God two miners while or two -6 -3 three along to minus long one one h 2 -1 And we got -2 plus long eight.

Okay, so for this integral, what we're gonna do is we're just going to split it up using the properties of integral. So we have the integral of a function where there isn't a term plus or minus another term. We can actually split that integral up into the integral of the first term, minus integral of the second term. So this is equal to the integral from negative 1 to 0 of two. X minus the integral of the X times dx. And then the next thing I'm gonna do is just take this two out of this first integral. Whenever we have a constant multiplied by our whole function and we're taking the integral, that function we can take the constant out and then multiply it by the integral of just the function. Okay, so now we have two times the integral of X dx. We're gonna be using this rule. I have written up in the top right to find the value of this integral. So X is the same as X to the first power. So here we would have an equaling one. So the integral of X. The first power is equal to X to the first plus one. Power, X squared divided by one plus one divided by two. So we're going to have two times X squared divided by two and then minus the integral of E. To the X is just itself. Eat to the X. And I should have had at least negative one to see us here. And so we're looking from negative 1-0. So in X is equal to zero. This first term is equal to zero. In the second term is equal to negative one. So we just get negative one when X is equal to zero And then the next is equal to -1. Could have minus Two times negative one squared is just too and then divided by two. I mean could actually just get rid of these. So I just kind of negative one squared which is one and then minus each negative first power, There's one divided by. And so this negative sign is going to distribute through this parentheses and we're going to get negative one minus one which is negative two and then plus one divided by E. And so this is the value of this definite integral.

Here we have the integral starting at zero. Going toe one of one plus one. Half you to before minus 2/5. You to the nine, do you? Let's go ahead and find the integral using. By undoing power rules, we have a U plus 1/10. You to the fifth, minus to over 50. You to the term quick from 01 Really? Here we can evaluate it one and also evaluated zero. And when we simplify, we get 53 over 50.


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