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Start the integration by parts process
**|udv = uv - |vdu**
u = (ln x)^{2} dv = x^{7} dx
du = 2(ln x)/x dx v = 1/8 x^{8}
= 1/8 x^{8} (ln x)^{2} - | 1/4(ln x)x^{7} dx
= 1/8 x^{8} (ln x)^{2} -1/4 | x^{7}(ln x) dx
Repeat the integration by parts method on the integral |x^{7}(ln x) dx
u=(ln x) dv = x^{7} dx
du = 1/x dx v = 1/8 x^{8}
= 1/8 (ln x) x^{8} - 1/8 | x^{7} dx
= 1/8 (ln x) x^{8} - 1/64 x^{8}
Simplify the answer (remebering to add the constant of integration).
= 1/8 x^{8} (ln x)^{2} -1/4 (1/8 (ln x) x^{8} - 1/64 x^{8 })
= **1/8 x**^{8} (ln x)^{2} -1/32 (ln x) x^{8} + 1/256 x^{8} + C

Answered by Rowan D.

Studies Astronomy and Phyiscs at Glasgow

Draw any right triangle with angle x in one of the corners. This is the key to properly understanding trigonometry: start by sketching the problem. Whenever you see cosine and sine, your first thought ought to be a triangle. Now, define the hypothenuse to be of length 1. The leg adjacent to the angle x will then have length cos(x), and the leg opposite the angle will have length sin(x) — this is one way cosine and sine are defined. Recall that the Pythagorean Theorem states that “a^{2} + b^{2} = c^{2}”, where a and b are legs and c is the hypothenuse. In this problem, we have “a=cos(x)”, “b=sin(x)”, and “c=1”. Now, we substitute those back into the Pythagorean Theorem to get “cos(x)^{2} + sin(x)^{2} = 1^{2} = 1”. Done.

To begin this question, we want to isolate the term containing the s, i.e. have the term containing the s by itself on one side of the equality. We do this by subtracting 8 from **both** sides. 4st = 5r - 8 To make s the subject of this equation, we want to **undo** whatever has been done to it. We notice that the s has been multiplied by 4 and by t, so, to counteract this, we *divide* both sides by 4 and t. s = (5r - 8) / 4t

Follow these steps: 1) Use the original price given, multiplying it by [(100-first discount percentage) over 100] *This is done to get the discounted price when the discount is applied once.* 2) Use the discounted price found in the previous step and multiply it by [(100-second disctount percentage) over 100] *This is done to find the final price of the product when the discount is applied twice.* For example, in a clothing department a type of cloth costs £100. The manager decides to reduce its price by 50% due to decrease in demand. After a month, the demand is still low so the manager decides to reduce its price by a further 20%. Following the steps mentioned above: 1) discounted price(after first discount) = 100 * [(100-50) over 100] = 100* 0.5 = £50 2) final price = 50 * [(100-20) over 100] = 50 * 0.8 = £40

Answered by Theofanis T.

Studies Mathematics at Southampton

1. Differentiating left hand side: 2(x+y)(1+dy/dx) from the chain rule
2. Differentiating right hand side: y^{2}+2xy(dy/dx) from the product rule
3. Equating sides and taking out factors of dy/dx to rearrange for dy/dx:
dy/dx=[y^{2}-2(x+y)]/[2(x+y)-2xy]
4. Substitute x=1 into original expression and solving for y (i.e. solving (1+y)^{2}=y^{2}) gives y=-1/2
5. Substituting x=1 and y=-1/2 into the expression for dy/dx gives dy/dx=-3/8

f(x) = 3x^3 + 2x^2 - 8x + 4
f(2) = 3(2)^3 + 2(2)^2 - 8(2) + 4
f(2) = 3(8) + 2(4) - 8(2) + 4
f(2) = 24 + 8 - 16 +4
f(2) = 20

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