HELP!! integration of trig identities (1 Viewer)

[star*eyed]

Please help with this question. Find the area bound by the curve y= sin x + cos x, the x-axis and the x = 0 and x= 3pi/2.

The answer is root 2 + 1 units^2. But could someone please help me with the working, cause obviously wat im doing is WRONG!

 

[Riviet]

You need to consider the parts of y=sinx+cosx that are below the graph, one of the parts below the x-axis is from x=3pi/4 to 7pi/4. So you need to split the integral into x=0 to x=3pi/4 and add it to the absolute value of the integral from x=3pi/4 to x=3pi/2.

Are you sure the answer's sqrt2 + 1?

 

[Raginsheep]

Notice that in the domain that you're given, the integral crosses the x-axis at 3pi/4.

 

[star*eyed]

sorry guys i wrote the wrong question, its is actually the area under the curve, the x axis, x=o and x = 3pi/4.

 

[Riviet]

Area=

3π/4aaaaaaaaaaaaaaaa3π/4
∫sinx+cosx dx = [sinx-cosx]
0aaaaaaaaaaaaaaaaaaaa0

= (1/rt2 + 1/rt2) - (0-1)

=2/rt2 + 1

=rt2 + 1 units²

 

[star*eyed]

Area=

3π/4aaaaaaaaaaaaaaaa3π/4
∫sinx+cosx dx = [sinx-cosx]
0aaaaaaaaaaaaaaaaaaaa0

= (1/rt2 + 1/rt2) - (0-1)

=2/rt2 + 1

=rt2 + 1 units²

how does the 2/rt2 suddenly just become rt2??

 

[Raginsheep]

2=rt2 * rt2

 

[Riviet]

how does the 2/rt2 suddenly just become rt2??

I rationalised the denominator:

.2 = a2(rt2) (by multiplying top & bottom by rt2)
rt2a(rt2)(rt2)

= 2(rt2)
aaa2

= rt2

 

[Rax]

how does the 2/rt2 suddenly just become rt2??

2/rt2 Think of it in index form

2¹ divided by 2^1/2 .

Because bases are the same becomes:

2^{1 - 1/2} .

Which is simply 2^1/2.

Hope that helps, thats the way I think of it anyway. Sometimes quicker than rationalising

 

[SoulSearcher]

That one only works when the bases are same.

 

[Rax]

Well yes, I was going to say that but I thought everyone would notice......
I have been suprised how many questions I have done which have the bases the same

Otherwise its good old fashion rationalising For the win