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# Calculus - WEEK 8 PARTICIPATION QUESTIONS

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WEEK 8 PARTICIPATION QUESTIONS...

1) Integration by substitution comes from Chain rule. Integration by parts is a consequence Product Rule of Derivatives. Prove that...

[f(x) g(x)]' = f(x) g' (x) + f(x) g(x)
Take the integral with respect to x of both sides of the equation, what will happen?

2) Retirement annuity. At age 25, Tom starts making annual deposits of \$2500 into an IRA account that pays interest at an annual rate of 5% compounded continuously. Assuming that his payments are made as a continuous income flow, how much money will be in his account if he retires at age 60? At age 65?

3) For positively defined functions the definite integral gives the area under the graph. Use the integral method to show that the area of the triangle can be obtained by multiplying Â½ of the base by the height.

https://brainmass.com/math/integrals/integration-by-substitution-using-chain-rule-242960

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WEEK 8 PARTICIPATION QUESTIONS...

1) Integration by substitution comes from Chain rule. Integration by parts is a consequence Product Rule of Derivatives. Prove that...

[f(x) g(x)]' = f(x) g' (x) + f(x) g(x)
Take the integral with respect to x of both sides of the equation, what will happen?

2) Retirement annuity. At age 25, Tom starts making annual deposits of \$2500 into an IRA account that pays interest at an annual rate of 5% compounded continuously. Assuming that his payments are made as a continuous income flow, how much money will be in his account if he retires at age 60? At age 65?

3) For positively defined functions the definite integral gives the area under the graph. Use the integral method to show that the area of the triangle can be obtained by multiplying Â½ of the base by the height.

(1) Given: [f(x) g(x)]' = f(x) g'(x) + f'(x) g(x)
Taking integrals with respect to x on both sides, we get
Integral [f(x) g(x)]' = Integral [f(x) g'(x) + f'(x) g(x)]
f(x) g(x) = Integral [f(x) g'(x)] + Integral [f'(x) g(x)]
Integral [f(x) g'(x)] = f(x) g(x) - Integral [f'(x) g(x)]
Let f(x) = u and g'(x) = dv, then we see that the above formula is equivalent to
Integral [u dv] = u v - Integral [v du], which is the formula for Integration by parts.

(2) FV = PMT * [(1 + i)^n - 1] / i, where i = e^r - 1
i = e^0.05 - 1 = 0.05127
(a) FV = 2500 * [(1 + 0.05127)^35 - 1] / 0.05127 = \$231841.36
(b) FV = 2500 * [(1 + 0.05127)^40 - 1] / 0.05127 = \$311539.70

(3) Let the equation of the straight line be y = mx + c. Let this line form a triangle together with the other two lines being
the x- axis and the y- axis.
x- intercept of the line is (put y = 0) x = -c/m
Area of the triangle = Integral [y dx] between the limits x = 0 and x = -c/m
= Integral [(mx + c) dx] between [0, -c/m]
= m(x^2)/2 + cx between [0, -c/m]
= m(c^2 / m^2)/2 + c(-c/m)
= -(1/2)(c^2 /m)
= -(1/2)(c/m)(c)
= -(1/2)(x- intercept)(y- intercept)
= (-1/2)(base)(height)
Area being an absolute value, we take Area = (1/2)(base)(height).

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