#$&*
course Mth 174
6/25 11
Calculus II Test 1Test should be printed using Internet Explorer. If printed from different browser check to be sure test items have not been cut off. If items are cut off then print in Landscape Mode (choose File, Print, click on Properties and check the box next to Landscape, etc.).
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• Numerical answers should be correct to 3 significant figures. You may round off given numerical information to a precision consistent with this standard.
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Test Problems:
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Problem Number 1
Find the general antiderivative of sin(x) / [ 7 + cos^ 5 (x) ].
Is this problem stated correctly? I couldn’t figure it out. I also tried to compute it on WolframAlpha and it gave me a very long answer. If it is right, I’m really confused.
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Letting u = cos(x) you get
-du / (1 + u^5).
This can be integrated, but it's actually beyond the scope of this course.
In this case I would give full credit just for doing the u substitution.
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Problem Number 2
Sketch a graph of a smooth curve which is asymptotic to y = 4 for positive values of x and to the y axis. Assuming that this graph represents the function f '', sketch graphs of f ' and f, assuming that both graphs pass through the origin.
f ` would have a curve concave down at the origin point. f would be a curve through the origin. It would decrease at a decreasing rate, then increase at an increasing rate.
Problem Number 3
Find the indefinite integral of the function t^ 5 cos(t^ 2).
Int (t^5 cos(t^2))=[=
u = t^2
du = 2t
1/2 du = t
1/2 int (u^2 cos(u) du)
f = u^2 g` = cos(u)
f` = 2u du g = sin(u)
1/2 u^2 sin(u) - 2 int (sin(u)u du)
u^2 sin(u) -2 int (sin(u)u du)
f = u g’ = sin(u)
f’ = du g = - cos(u)
1/2 u^2 sin(u) - 2u sin(u) - int (cos(u)du)
1/2 u^2 sin(u) -2 u sin(u) + sin(u) + c
1/2 t^4 sin(t^2) - 2 t^2 sin(t^2) + sin(t^2)
Note: Is this the best way to find the solution?
@&
Yes. u^2 cos(u) is easily enough integrated with two integrations by parts, as you have done.
However check your solution to make sure its derivative is indeed u^2 cos(u).
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Problem Number 4
The rate at which water rises in a container is r(t) cm / sec, where t is clock time in seconds.
• Write an expression for the change in water depth between clock times t and t + `dt.
• Write a Riemann sum with n intervals expressing the change in water depth between t = 3 and t = 18.
• If r(t) = .7 t^2 + 3 t + 78, then how much did depth change between t = 3 and t = 18?
• If the depth was initially 3700, then what is an expression for y(t), the depth at clock time t?
I’m not exactly sure how to solve this one.
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The rate at which water rises is r(t), the time interval is `dt, so the rise will be r(t) `dt.
The Riemann sum will involve terms of the form r(t_i) * `dt.
The Riemann sum represents the approximate total change in depth.
As the number of intervals approaches infinity, the Riemann sum will approach an integral, which gives you the exact change in depth.
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Problem Number 5
Integrate the function f(x) = 6 cos( 3 x) from x = .7 to x = 1.5.
int(6 cos(3x)dx)
u = 3x
du = 3
2 du = 6
2 int(cos(u)du)
2sin(u) = 2sin(3x)
2sin(3 * 1.5) - 2sin(3 * .7) = -3.681
Problem Number 6
Find the integral of t sin( 3 t^2) between t = 1.5 and t = 2.7 in two ways.
• First find an antiderivative of the function, in terms of the original variable t, and apply the First Fundamental Theorem.
• Then use an appropriate u = ... substitution and rewrite the integral in terms of u. Don't convert the antiderivative back to the original variable, but simply apply the First Fundamental Theorem to an antiderivative expressed in terms of u.
int(tsin(3t^2)dx)
u = 3t^2
du = 6t
1/6 int(sin(u) du)
1/6 -cos(3t^2) + c
1/6 -cos (3(2.7)^2) - 1/6 -cos(3(1.5)^2) = 1.885
Note: I’m not sure what the first part is asking. I’m not sure how to solve this without u substitution.
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You used u substitution to get the integral in terms of t, which is what you need to do.
However the integral of 1/6 sin(u) du is -1/6 cos(u), not 1/6 - cos(u).
If t changes from 1.5 to 2.7, what are the corresponding values of u?
What is the definite integral of 1/6 cos(u) between these limits?
What is the expression for the integral oft sin(3t^2) dt, and what is the definite integral from t = 1.5 to t = 2.7?
How do the two definite integrals compare?
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Problem Number 8
Find the general solution of dy / dx = 3 e^x - 8.
dy / dx = 3 e^x - 8
= 3e^x - 8x + c
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If you mean
dy / dx = 3 e^x - 8
y = 3e^x - 8x + c
then you are correct.
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