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Calculus  II                                                          Test #1                                                  March 5, 2003

Name____________________                     R.  Hammack                                                Score ______
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(1) Find the following integrals.

(a) ∫ (2x^3 + 1) dx =x^4/2 + x + C

(b) ∫x x^(1/2) dx =∫x x^(1/2) dx =∫x^(3/2) dx =x^(5/2)/(5/2) + C= (2x^(5/2))/5 + C= (2x^(1/2)^5)/5 + C

(c) ∫ (4e^x + 1/x + sec(x) tan(x) ) dx =4e^x + ln | x | +sec(x) + C

(d) ∫1/(x^2 + 1) dx =tan^(-1) (x) + C

(e) ∫ (2x)/(x^2 + 1) dx =∫1/udu = ln | u | +C = ln | x^2 + 1 | +C

u = x^2 + 1
d u = 2x d x

(f) ∫ (3x + 2)^7dx =1/3∫ (3x + 2)^73dx =1/3∫u^7du =1/3u^8/8 + C= (3x + 2)^8/24 + C

u = 3x + 2
d u = 3 d x

(2) Find the following definite integrals.

(a)   ∫__ (-1)^1 (2x + 1) dx =[x^2 + x] _ (-1)^1 = (1 + 1) - (1 - 1) = 2

(b)    ∫__ (-π)^0cos(x) dx =[sin(x)] _ (-π)^0 = sin(0) - sin(π) = 0

(c)  ∫_ln(3)^ln(4) e^x/(e^x + 4) dx =∫_ (e^ln(3) + 4)^(e^ln(4) + 4) 1/udu =∫_7^81/udu =
[ln(u)] _7^8 = ln(8) - ln(7) = ln(8/7)

u = e^x + 4
d u = e^x d x


(d)  ∫__0^1 (2x - 1)^4dx =1/2∫__ (2 (0) - 1)^(2 (1) - 1) (2x - 1)^42dx =1/2∫__ (-1)^1u^4du = 1/2[x^5/5] _0^1=1/5

u = 2x - 1
d u = 2 d x


(3)  The expression   Underscript[lim , n∞](Underoverscript[∑ , k = 1, arg3] x_k ^* cos(x_k ^*)^(1/2) Δx )   represents a definite integral over the interval [3, 5].   Write the definite integral. (You do not need to find its value.)   

  ∫__3^5x cos(x)^(1/2) dx

(4)  Find the average value of f(x) = x^2 on the interval [0, 3].

1/(3 - 0) ∫__0^3x^2dx = 1/3[x^3/3] _0^3 = 3




(5) Find the following integrals. You may find it easiest to consider the area under the graphs.

(a) ∫_ (-4)^0 (16 - x^2)^(1/2) dx = (π 4^2)/4 = 4π
Region is one fourth of a circle of radius 4.


(b) ∫_ (-2)^4 | x - 2 | dx =1/2(4)(4) + 1/2(2)(2) = 10.
Region is two triangles.


(6) Find the derivative of the function F(x) =    ∫__3^x^4 (ln(t) + t  )^(1/2) dt

F ' (x) =    (ln(x^4) + x^4 )^(1/2) 4x^3


(7) A  train, moving with constant acceleration, travels 25 miles in half an hour.  At the beginning of the half-hour period, it has a velocity of 10 miles per hour.  What is its velocity at the end of the half-hour period?

The information says:
s(0) = 0
s(1/2) = 25
v(0) = 10
Let a be the constant acceleration.

Know: v(t) = ∫a dt = a t + C
Then 10 = v(0) = a(0) + C, meaning C = 10.
Thus v(t) = a t + 10.

If we could just find a, then we would have the formula for velocity, and the answer to the problem would be v(1/2). The  information that we have not used yet is s(0) = 0 and s(1/2) = 25. That is information about position, so to use it we must make the position function.

Know: s(t) = ∫v(t) dt = ∫ (a t + 10) dt = (a t^2)/2 + 10t + C
Then 0 = s(0) = (a 0^2)/2 + 10 (0) + C, meaning C = 0.
Thus s(t) = (a t^2)/2 + 10t
Now, 25 =s(1/2) = a(1/2)^2/2 + 10 (1/2)
So  25 =a/8 + 5
And  20 =a/8
So a = 160

Finally, we now have the velocity function as v(t) = 160t + 10.

The velocity at the end of the half-hour period is v(1/2) = 160(1/2) + 10 = 90mph.



(8) Suppose that    ∫_3^2f(x) dx = 3  and  ∫_2^6f(x) dx = 9.

(a)    ∫_2^35f(x) dx = -5∫_3^25f(x) dx = -(5) (3) = -15


(b)    ∫_3^6f(x) dx = 12, as follows:

∫_2^6f(x) dx = ∫_2^3f(x) dx + ∫_3^6f(x) dx

∫_2^6f(x) dx = -∫_3^2f(x) dx + ∫_3^6f(x) dx

9 = -3 + ∫_3^6f(x) dx

∫_3^6f(x) dx = 12