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# for the definate integral :

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for the definate integral :
Upper 2 lower -1, of 5sin(3x)dx
=5[(-cos3(2)/3)-(-cos3(-1))/3]
=5[(-cos(6)/3)+(cos(-3))/3]
=5[(-cos(6)/3)-(cos(3))/3]

Answer to find (if right in the sample exam is equal to -3.25, which is what a graphing calculate gives.

Also, how do I put the answer in manually into a calculator that does not graph, that is, where do i put brackets/ what steps do i take to get the right answer. Nothing I do can provide the right answer. Yes calculator is in degrees.

Jun 1, 2014

#6
+5

ok.  So plot the graph of cos(x) where x is is in degrees.  This time we have to home in on a small portion of the graph to see it clearly: You can see they are still the same.

$$\underset{\,\,\,\,^{{360^\circ}}}{{cos}}{\left({\mathtt{3}}^\circ\right)} = {\mathtt{0.998\: \!629\: \!534\: \!755}}$$

$$\underset{\,\,\,\,^{{360^\circ}}}{{cos}}{\left(-{\mathtt{3}}^\circ\right)} = {\mathtt{0.998\: \!629\: \!534\: \!755}}$$

However, your original post stated that the correct answer to the integral was -3.25.  This is only true if x is in radians.

Jun 1, 2014

#1
+5

Wasn't this the one you did last night?? Remember the calculator has to be in radian mode.......

After you integrate, you should get

(5/3) [cos(-3) - cos(6)] = -3.25

Does that help??   Jun 1, 2014
#2
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You were almost right Stu; just one small (but vital) mistake in the last line:  cos(-3) is +cos(3) not -cos(3).

Jun 1, 2014
#3
0

Why is this in radians and not degrees? It is not specified so I assume it is degree standard. Also, I do not understand how you take the /3 out side the bracked and devide the 5 to get the same result as having it inside. Also, why is it + when there is 3 - sign in the equation. I was told that the two intial minu make positive but then the - on the degree to the cos (-3) comes out side to make the + and - a negative. As shown in my example. SOme further explaination would assist me. As well, this is not the equation from last night.

Jun 1, 2014
#4
+5

It is normal to assume the angle is in radians unless it is accompanied by an explicit degree sign.

To see that cos(-3) is the same as cos(3) look at the graph below: You can see that cos(-3) has exactly the same value as cos(3).  They are both ≈ -0.99.

Jun 1, 2014
#5
0

im not sure, but only one question is in radian on the exam sample, which specifies radians and gives the sign of radians. All others I have taken to be in degrees.

Jun 1, 2014
#6
+5

ok.  So plot the graph of cos(x) where x is is in degrees.  This time we have to home in on a small portion of the graph to see it clearly: You can see they are still the same.

$$\underset{\,\,\,\,^{{360^\circ}}}{{cos}}{\left({\mathtt{3}}^\circ\right)} = {\mathtt{0.998\: \!629\: \!534\: \!755}}$$

$$\underset{\,\,\,\,^{{360^\circ}}}{{cos}}{\left(-{\mathtt{3}}^\circ\right)} = {\mathtt{0.998\: \!629\: \!534\: \!755}}$$

However, your original post stated that the correct answer to the integral was -3.25.  This is only true if x is in radians.

Alan Jun 1, 2014
#7
0

Thanks, I have emailed the guy who wrote the exam to clarify facts about this matter. Thanks.  In other questions  of indefinate integrals, I believe it was right when calculated in degrees. But have had some problems with this so im not sure. I'l wait for a reply email from the person in question. Thanks for this assistance.

Jun 1, 2014