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Day 42 - Test #5 Review Packet
3) Because neither of the factors of the denominator divides into the numerator evenly, you won’t get “0” for a remainder in synthetic division. This is why you need to use polynomial long division. Here’s how that works:
i) Write the denominator on the outside of the division sign, and the numerator on the inside of the division sign.
ii) Figure out what the first term of the outside needs to be multiplied by to equal the first term of the inside.
x^2 times what will equal x^3?
It would be x. Write x above the division sign, right above the x term on the inside.
iii) Multiply the outside equation by x, and write the answer underneath the inside of the division sign.
(x^2 - 2x - 8) * x = x^3 - 2x^2 - 8x
iv) Subtract this new expression from the inside of the division sign.
(x^3 - 7x^2 + 5x) - ( x^3 - 2x^2 - 8x) = -5x^2 + 13x
v) Drop down the next term (+40) and repeat steps #2-4.
x^2 times what will equal -5x^2? …it’s -5. Write this above the division sign.
Multiply: (x^2 - 2x - 8) * -5 = -5x^2 + 10x + 40
Subtract: (-5x^2 + 13x + 40) - (-5x^2 + 10x + 40) = 3x
vi) Your final remainder is 3x, which means this is the only term that can’t be divided evenly by your original denominator. Your answer is the expression from above the division sign, plus the remainder / original denominator.
Answer to this part: x - 5 + 3x / (x^2 - 2x - 8)
Integrate x and -5 like a normal integral, and integrate 3x / (x^2 - 2x - 8) by using the partial fraction shortcut.
6) You can actually write either a tan^-1 or cot^-1 equation to begin this question, and you should get the same derivative answer either way.
8a) The anti-derivative you should get is -1 / 3x^3. So, when you plug in the bounds, it looks like:
-1 / 3a^3 - -1 / 3*3^3
If a —> infinity, then the denominator of the first term is getting really big forever, which means that fraction is getting very close to zero. So, its limit is zero. The second fraction is equal to 1/81, so you end up with 0 + 1/81. Because you get an actual numerical answer (not infinity for a final answer), that means it converges.
8d) The reason you split it into two halves is because the integral isn’t continuous. You can only use FTC on a continuous area (this is one of the reasons that improper integrals exist!). So, we split it up at the point of discontinuity, and integrate both parts of the integral area separately.
Technically, you’d still get the right answer if you never split it into two halves. But, the math you’d be doing would be incorrect, because you’d be saying that you’re finding the area under a curve that’s discontinuous. You can’t do that.
3) Because neither of the factors of the denominator divides into the numerator evenly, you won’t get “0” for a remainder in synthetic division. This is why you need to use polynomial long division. Here’s how that works:
i) Write the denominator on the outside of the division sign, and the numerator on the inside of the division sign.
ii) Figure out what the first term of the outside needs to be multiplied by to equal the first term of the inside.
x^2 times what will equal x^3?
It would be x. Write x above the division sign, right above the x term on the inside.
iii) Multiply the outside equation by x, and write the answer underneath the inside of the division sign.
(x^2 - 2x - 8) * x = x^3 - 2x^2 - 8x
iv) Subtract this new expression from the inside of the division sign.
(x^3 - 7x^2 + 5x) - ( x^3 - 2x^2 - 8x) = -5x^2 + 13x
v) Drop down the next term (+40) and repeat steps #2-4.
x^2 times what will equal -5x^2? …it’s -5. Write this above the division sign.
Multiply: (x^2 - 2x - 8) * -5 = -5x^2 + 10x + 40
Subtract: (-5x^2 + 13x + 40) - (-5x^2 + 10x + 40) = 3x
vi) Your final remainder is 3x, which means this is the only term that can’t be divided evenly by your original denominator. Your answer is the expression from above the division sign, plus the remainder / original denominator.
Answer to this part: x - 5 + 3x / (x^2 - 2x - 8)
Integrate x and -5 like a normal integral, and integrate 3x / (x^2 - 2x - 8) by using the partial fraction shortcut.
6) You can actually write either a tan^-1 or cot^-1 equation to begin this question, and you should get the same derivative answer either way.
8a) The anti-derivative you should get is -1 / 3x^3. So, when you plug in the bounds, it looks like:
-1 / 3a^3 - -1 / 3*3^3
If a —> infinity, then the denominator of the first term is getting really big forever, which means that fraction is getting very close to zero. So, its limit is zero. The second fraction is equal to 1/81, so you end up with 0 + 1/81. Because you get an actual numerical answer (not infinity for a final answer), that means it converges.
8d) The reason you split it into two halves is because the integral isn’t continuous. You can only use FTC on a continuous area (this is one of the reasons that improper integrals exist!). So, we split it up at the point of discontinuity, and integrate both parts of the integral area separately.
Technically, you’d still get the right answer if you never split it into two halves. But, the math you’d be doing would be incorrect, because you’d be saying that you’re finding the area under a curve that’s discontinuous. You can’t do that.
