The area between y = x^2-4x+10 and y = 4x-x^2 over [1,3] is 4/3 . 10/3 . 16/3 .

Question

The area between y = x^2-4x+10 and y = 4x-x^2 over [1,3] is

4/3 .

10/3 .

16/3 .

none of the above.

The average value of f(t) = t^3 on [0,a] (a > 0) is overline (f)_([0,a]) = (a/2)^2. Find a.

a = 1.

a = 2.

a = 3.

The volume of the solid obtained by rotating about the y axis the region bounded between by y = 4-(x-2)^2 and the x-axis is expressed by the integral:

pi integral from 0^3[4-(x-2)^2]dx

2 pi integral from 0^3[4-(x-2)^2]dx

pi integral from 0^3[4-(x-2)^2]^2dx

none of the above.

Let l(x) denote the arc length of y =-root (1-t^2) on t in [-1,x]. Then

l(0) = 0.

l(0) = pi/4 or l(0) = pi/2.

l(0) = integral from (-1)^0 root (1+2t) dt.

l(0) = 1.

A force of 4 N will stretch a rubber band 0.04m. Assuming that Hooke's Law applies, how much work does it take to stretch the rubber band by 10 cm?

10 J

20 J

15 J

none of the above.

Explanation / Answer

1)

(A=int_{1}^{3}(x^2-4x+10)-(4x-x^2)dx=int_{1}^{3}2x^2-8x+10dx=52/3-32+20=16/3)

So (3)

2)

(Avg.=int_{0}^{a}f(t)dt/a=int_{0}^{a}t^3dt/a=a^3/4=a^2/4)

So

a=0 or 1

So (1)

3)NOne (4)

4)

(L(0)=int_{-1}^{0}sqrt{1+(dy/dt)^2}dt=int_{-1}^{0}sqrt{1+(t^2/(1-t^2))}dt=int_{-1}^{0}1/sqrt{(1-t^2}dt=pi/2)

So (2)

5)k*0.04=4

So k=100N/m

So

For x=0.1m

W=0.5kx^2=0.5*100*0.01=0.5J

So (4)

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