Calculus Midterm #1

Calculus Midterm #1

Spring 2005

(1) Find the following integrals.

(i)

₇

0 x√³

x+ 1dx. 10%

(ii)

x(lnx)²dx. 10%

(iii)

xe^2xdx. 10%

(2) The volume of the solid formed by revolving the graph off(x) about thex-axis 10%

is given by the formula:

1

0 π(f(x))²dx.

Find the volume by revolving f(x) = _x¹,1≤x <∞ about thex-axis.

(3) After test-marketing a new menu item, a fast-food restaurant predicts that sales 10%

of a new item will grow according to the model dS

dt = 2t (t+ 4)²

where t is the time in weeks and S is the sales in thousands of dollars. Find the sales of the menu item at 10 weeks.

(4) The Simpson Rule (for even n) approximates

_b

a f(x)dx by 10%

(b−a

3n )[f(x0) + 4f(x1) + 2f(x2) +· · ·+ 4f(xn−1) +f(xn)]. Use n = 6 to estimate

₁

−1

√1 2πe^−x

2

2 dx up to three digits after the decimal point. (Note: ^√1_2π = 0. .399)

(5) Find the second partial derivatives fxx and fxy of f(x, y) =e^−2y/x. 10%

(6) Find and sketch the domain of the function 5%

f(x, y) =

4−x²−y².

(7) A company’s production is given by the Cobb-Douglas production function Q=f(x, y) = 4x^0.5y^0.5

where x > 0 is the number of units of labor and y >0 is the number of units of capital.

(i) What is the production level whenx= 100 and y = 144. 5%

(ii) Sketch the level curves for Q = 4,6,8,10,12. On your drawing, show the 5%

coordinates of the points at which the level curves intersect with y= 1.

Solutions for Calculus Midterm 1 (1) (i) Letu=x+ 1, then du=dx.

And we have Z ⁷

0

x√³

x+ 1dx= Z ⁸

1

(u−1)u¹³du

= Z ⁸

1

(u⁴³ −u¹³)du

= 3

7u⁷³ −3 4u⁴³

8

1

= 1209 28 . (ii) Letu= (lnx)², dv=xdx.

Then du= _x²lnxdx, v = ¹₂x². Thus R

x(lnx)²dx = ¹2x²(lnx)²−R

xlnxdx obtained from using integration by parts.

Letu= lnx, dv=xdx⇒du= ¹_xdx,v = ¹₂x². Then we use integration by parts again.

Thus Z

x(lnx)²dx= 1

2x²(lnx)²−(1

2x²lnx− Z 1

2xdx)

= 1

2x²(lnx)²− 1

2x²lnx+1

4x²+C

= 1

2x²[(lnx)²−lnx+ 1 2] +C (iii) Letu=x, dv=e^2xdx⇒du=dx,v = ¹2e^2x

Then we apply integration by parts and obtain Z

xe^2xdx = 1

2xe^2x− Z 1

2e^2xdx

= 1

2xe^2x− 1

4e^2x+C

(2) Z ^∞

1

π(1 x)²dx

= lim

b→∞

Z b

1

π 1 x²dx

= lim

b→∞

[−π x]

b

1

= lim

b→∞

[−π b +π]

=π

Thus the volume is π cubic units.

(3) dS

dt = 2t (t+ 4)².

And we know that S(0) = 0.

Thus S(10)−S(0) = Z ¹⁰

0

2t

(t+ 4)²dtby Fundamental Theorem of Cal- culus. It follows that S(10) =

Z ¹⁰

0

2t (t+ 4)²dt. Let u=t+ 4⇒du=dt

Then

S(10) = Z ¹⁴

4

2(u−4) u² du

= Z ¹⁴

4

(2 u − 8

u²)du

= 2 lnu+ 8 u

14

4

= (2 ln 14 + 8

14)−(2 ln 4 + 2)

= 2 ln 7−2 ln 2− 10 7

≈1.0769

Hence the sales of the menu item at 10 weeks are 1.0769 thousand dollars.

(4) n = 6

⇒x0 =−1,x1 =−²³, x2 =−¹³, x3 = 0, x4 = ¹3,x5 = ²3, x6 = 1.

Then by Simpson’s Rule, we have Z ¹

−1

√1 2πe^−x

2

2 dx= 1

√2π 2

18(e⁻¹² + 4e⁻⁹² + 2e⁻¹⁸¹ + 4e⁰+ 2e⁻¹⁸¹ + 4e⁻²⁹ +e⁻¹²)

≈0.399×1

9(0.6065 + 4×0.8007 + 2×0.9460 + 4×1 + 2×0.9460 + 4×0.8007 + 0.6065)

≈0.683 (5)

f(x, y) =e^−x2y f_x =e^−x2y2yx⁻²

⇒f_xx = ∂

∂x(e^−x2y)2yx+e^−x2y ∂

∂x(2yx⁻²)

=e^−x2y2yx⁻²(2yx⁻²) +e^−x2y(−4yx⁻³)

= 4x⁻⁴ye^−x2y(y−x) And f_xy = ∂

∂y(e^−x2y)2yx⁻²+e^−x2y ∂

∂y(2yx⁻²)

=e^−x2y(−2

x )2yx⁻²+e^−x2y(2x⁻²)

= 2x⁻³e^−x2y(x−2y)

(6) f(x, y) =p

4−x²−y²

4−x²−y² ≥0⇒x²+y² ≤4.

Thus the domain of f is the set of all points that lie on or inside the circle given byx²+y² = 2². That is dom(f) = {(x, y)∈R²|x²+y² ≤4}. And we sketch the domain as shown in the following figure 1.

2 2

−2

−2 x

y

Figure 1: The domain off(x, y).

(7) (i) The production level is Q=f(100,144) = 4(100)^0.5(144)^0.5 = 480 units.

(ii) Q= 4⇒xy = 1 Q= 6⇒xy = ⁹₄ Q= 8⇒xy = 4 Q= 10⇒xy = ²⁵₄ Q= 12⇒xy = 9

Hence we obtain five level curves and show it in the figure 2.

Furthermore, these level curves intersect with y = 1 at points A = (1,1), B = (⁹₄,1), C = (4,1), D = (²⁵₄ ,1), and E = (9,1) respectively.

E D

C B

A

0 2 4 6 8 10

y

2 4 6 8 10

x

Figure 2: The level curves for Q=4,6,8,10,12.

(iii) When the capital is fixed at y = 1, from the figure 2 we can find that the production level Q increases 2 units, the distances AB, BC, CD, and DE become larger. Therefore the surface is less steep.

(8) f(x, y) = 12y+ 6x−x²−y³ f_x = 6−2x,f_y = 12−3y²

We solve f_x = 0 andf_y = 0, then determine that the critical points are (3,2) and (3,−2).

Furthermore f_xx =−2, f_yy =−6y, f_xy = 0.

And we use the second-partials test to classify the critical points as follows.

1. When (x, y) = (3,2),

2. When (x, y) = (3,−2),

d=f_xx(3,−2)f_yy(3,−2)−[f_xy(3,−2)]² = (−2)12−0 =−24<0.

Thus (3,−2) is a saddle point.