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MATH 105 PRACTICE MIDTERM III Key

Please work out each of the given problems. Credit will be based on the steps that you show towards the final answer. Show your work.

PROBLEM 1 Please answer the following true or false. If false, explain why or provide a counter example. If true, explain why.

A) If f(x) is a positive continuous function such that

$ \int_1^3 f(x)dx = 4 $

then

$ \int_0^5 f(x)dx$

cannot be equal to 3.

Solution

True, since f is a positive function, both integrals represent areas, and the second integral represents an area of a region that contains the region of the first integral. Hence the second integral must be at least as large as 4.

B) If f(x) is a differentiable function such that the equation of the tangent line at

x = 2

                   1           1
        y =         x -
                   2           2

and if x = 2 is the first guess in Newton’s method, then x = 1 is the second guess.

Solution

True, We arrive at the second guess by finding the x-intercept of the tangent line. Since the x-intercept is 1, x = 1 is the second guess.

C) If f(x) and g(x) are continuous functions on [a,b], then

$ \int_a^b f(x)g(x)dx =(\int_a^b f(x)dx) (\int_a^b g(x)dx) $

Solution

False, for example, if f(x) = g(x) = 1 and if a = 0 and b = 2, then the left hand side is 2 and the right hand side is (2)(2) = 4

PROBLEM 2 Evaluate the following integrals:

A. $ \int (sin(x) - \sqrt{x} + 2x^2 - 3)dx$

Solution

Just integrate the terms individually:

-cos x - 2/3 x^3/2 + 2/3 x³ - 3x + C

B. $ \int_2^3 \frac{x}{(1 - x)^3}dx$
Solution

We use u-substitution:

u = 1 - x du = -1dx dx = -du x = 1 - u

when

x = 2 u = -1

when

x = 3 u = -2

Substituting produces

$ -\int_{-1}^{-2} \frac{1-u}{u^3}du = -\int_{-1}^{-2}(u^{-3}-u^{-2})du $

$ = -[\frac{u^{-2}}{-2} - \frac{u^{-1}}{-1}]_{-1}^{-2} = -[(\frac{(-2)^{-2}}{-2} - \frac{(-2)^{-1}}{-1} ) - (\frac{(-1)^{-2}}{-2} - \frac{(-1)^{-1}}{-1} )] $

$ = -[(-\frac{1}{8} + \frac{1}{2}) - (-\frac{1}{2}+1)] = -[\frac{3}{8} + \frac{1}{2}] = -\frac{7}{8} $

C. $ \int csc(2x)cot(2x)dx$
Solution

We use u-substitution:

u = 2x du = 2dx dx = 1/2 du

Substituting, we get

$ \frac{1}{2} \int csc(u) cot(u) du = - \frac{1}{2} csc(u) + C $

Resubstituting, gives

-1/2csc(2x) + C

PROBLEM 3 Use Riemann Sums to find the area of the region below the curve y = 9 - x², above the x-axis, and between x = 1 and x = 3.

Solution

We have

Dx = (3 - 1) / n = 2/n

Hence

$ \text{Area} = \lim\limits_{n \to \infty} \sum_{i=1}^{n} [9 - (1 + \frac{2i}{n})^2 ]\frac{2}{n} $

$ = \lim\limits_{n \to \infty} \sum_{i=1}^{n} [9 - (1 + \frac{4i}{n} + \frac{4i^2}{n^2} ) ]\frac{2}{n} $

$ = \lim\limits_{n \to \infty} \sum_{i=1}^{n} [ \frac{16}{n} - \frac{8i}{n^2} - \frac{8i^2}{n^3} ] $

$ = \lim\limits_{n \to \infty} [ \frac{16}{n}\sum_{i=1}^{n} 1 - \frac{8}{n^2}\sum_{i=1}^{n} i - \frac{8}{n^3}\sum_{i=1}^{n} i^2 ] $

$ = \lim\limits_{n \to \infty} [ \frac{16n}{n} - \frac{8n(n+1)}{2n^2} - \frac{8n(n+1)(2n+1)}{6n^3} ] $

$ = \lim\limits_{n \to \infty} [ 16 - \frac{4n+4}{n} - \frac{8n^2 + 12n + 4}{3n^2}] $

$ = 16 - 4 - \frac{8}{3} = \frac{28}{3} $

PROBLEM 4

Let

$ \int_3^{sin(x)} cos(t^2)dt$

Find F'(x)

Solution

Use the chain rule along with the second fundamental theorem of calculus:

u = sin x $ F(u) = \int_3^u cos(t^2)dt $

u'(x) = cos x

F'(u) = cos u² = cos(sin² x) By the fundamental theorem of calculus

Now use the chain rule

F'(x) = u'(x)F'(u) = (cos x)(cos(sin²x))

PROBLEM 5

You are the owner of Tahoe Winter Wear and need to determine the best price to sell your most popular winter jacket. Your cost for selling the jackets is

C = 50 + 20x

Where x is the amount to jackets that you sell. Your research shows that the relationship between price, p, and the number of jackets that you can sell, x, is
p = 300 – 10x
How much should you charge for your jacket in order to maximize profit?

Solution

First calculate the revenue R:

R = xp = x(300 - 10x) = 300x - 10x²

Now use the fact that Profit equals Revenue minus Cost:

P = R - C = (300x - 10x²) - (50 + 20x) = 280x - 10x² - 50

To find the maximum profit we take the derivative and set it equal to zero:

P' = 280 - 20x = 0

x = 14

Now substitute 14 for x in the demand equation:

p = 300 - 10(14) = 160

You should charge $160 for the jacket.

PROBLEM 6

You are manufacturing a square computer chip. Your machine can construct the square with side length 0.4 0.0002 cm. Use differentials to approximate the maximum percent error in the area of the chip.

Solution

Use the formula

A = x²

A' = 2x

Differentials gives

DA @ 2xDx = 2(0.4)(0.0002) = 0.00016

To find the percent error, use

Percent Error = (DA/A)(100%) = (0.00016/.4²)(100%) = 0.1%

PROBLEM 7

Let f(x) = x³ + x + 4

Prove that f(x) has an inverse function.
Solution
We use the theorem that tell us that if f(x) is monotonic then the inverse exists. We have

f '(x) = 3x² + 1

Which is always positive, hence f(x) is monotonically increasing. Therefore f has an inverse.
Let g(x) be the inverse of f(x). Find g'(4) .
Solution
We use the formula
1

g '(4) =
                          f '(g(4))

We find g(4) by setting f(x) equal to 4:

        4 = x³ + x + 4

        x³ + x = 0            x(x² + 1) = 0

Hence x = 0. Now calculate

        f '(0) = 3(0)² + 1 = 1

Finally

                        1
        g '(4) =           = 1
                         1

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