Integration Q: If k = int, -2 to 3, of sin ( x^(1/3)), ....

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grapz

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If k = [int] from -2 to 3 of sin ( x^(1/3))

what is k?

The choices given are

k> 3
2<k <=3
0<k<=1
k<= 0

Note, this question is intended to be done with out a calculator. Because the integral is from -2 -->3 it is hard to figure out what the exact value is so how does one finish this problem?
 
Re: Integration Question

Sin is an odd function - so integration from -2 to +2 would be zero. [edit]

2[sup:21ttttri]1/3[/sup:21ttttri] and 3[sup:21ttttri]1/3[/sup:21ttttri] are very near to \(\displaystyle \frac {\pi}{2}\) --- does that give any hint about the integration (area under the curve)??
 
Re: Integration Question

Use numerical integration or a Taylor series. Either way, enjoy the grunt work (-2 to 3)?. Have fun

.
 
Re: Integration Question

grapz said:
If k = [int] from -2 to 3 of sin ( x^(1/3))

what is k?

The choices given are

k> 3
2<k <=3
0<k<=1
k<= 0

Note, this question is intended to be done with out a calculator. Because the integral is from -2 -->3 it is hard to figure out what the exact value is so how does one finish this problem?
Click to expand...

You only need to "estimate" the answer here.

The domain of integration reduces to +2 -> +3 (because -2 -> +2 integrates to 0)

Sine is always positive in the given domain (2 to 3)

Value of sine is <= 1

So what should be the "restriction" on value of k?
 
Note: integral[sin(x^(1/3)),x] = -3x^(2/3)cos(x^(1/3)) + 6cos(x^(1/3)) + 6x^(1/3)sin(x^(1/3)) + C

Can be done by integration by parts.

Integral[sin(x^(1/3)),x,-2,3] = about .975499218424 (this I got off my TI-89).
 
My mistake, didn't read on, as I see only an estimate.was needed.

By the way, very good analysis Subhotosh Khan, as I should have known something was awry.
 
The official solution to this probelm was that first you realize its odd function so the integral is basically only the integral from 2-> 3 of sin (x^1/3) dx.

This is the part i don't understand. I was hoping someone can explain it to me.

0 < sin ( x^1/3) <= 1. Hence 0 < int from 2--> 3 sin(x^1/3) dx <= 1 And hence 0 < k <= 1

I know the sin function is from 0 to <= 1 but how can you assume that the integral of that function from 2--> 3 is also between 0 and 1?
 
Suppose we have two integrable functions.
If \(\displaystyle x \in \left[ {a,b} \right]\left[ {0 \leqslant g(x) \leqslant f(x)} \right]\) then it is true that \(\displaystyle \int\limits_a^b {g(x)dx} \leqslant \int\limits_a^b {f(x)dx}\).

So on [2,3] we have \(\displaystyle 0 \leqslant \sin \left( {\sqrt[3]{x}} \right) \leqslant 1\quad \Rightarrow \quad \int\limits_2^3 {\sin \left( {\sqrt[3]{x}} \right)dx} \leqslant \int\limits_2^3 {1dx} = 1\)
 
grapz said:
The official solution to this probelm was that first you realize its odd function so the integral is basically only the integral from 2-> 3 of sin (x^1/3) dx.

This is the part i don't understand. I was hoping someone can explain it to me.

0 < sin ( x^1/3) <= 1. Hence 0 < int from 2--> 3 sin(x^1/3) dx <= 1 And hence 0 < k <= 1

I know the sin function is from 0 to <= 1 but how can you assume that the integral of that function from 2--> 3 is also between 0 and 1?
Click to expand...

Do you know that - the integration is basically the area under the curve - for the given domain?

If so then height of the area does not go over 1 (f(x)<= 1) - and the base = 1 (= 3-2).

Then the area must be (height x base = ) <=1.
 
Ah yes thank you, i forgot that the integral was from 2--> 3 for a moment
 
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