# How do you find the average value of the function for #f(x)=sinxcosx, 0<=x<=pi/2#?

The average value is

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To find the average value of the function ( f(x) = \sin(x)\cos(x) ) over the interval ( 0 \leq x \leq \frac{\pi}{2} ), you can use the formula:

[ \text{Average value} = \frac{1}{b - a} \int_a^b f(x) , dx ]

where ( a ) and ( b ) are the limits of integration. In this case, ( a = 0 ) and ( b = \frac{\pi}{2} ). So, we have:

[ \text{Average value} = \frac{1}{\frac{\pi}{2} - 0} \int_0^{\frac{\pi}{2}} \sin(x)\cos(x) , dx ]

Now, integrate ( \sin(x)\cos(x) ) over the interval ( 0 \leq x \leq \frac{\pi}{2} ):

[ \int_0^{\frac{\pi}{2}} \sin(x)\cos(x) , dx = \frac{1}{2} \int_0^{\frac{\pi}{2}} \sin(2x) , dx ]

Using the identity ( \sin(2x) = 2\sin(x)\cos(x) ), we get:

[ = \frac{1}{2} \left[ -\frac{1}{2} \cos(2x) \right]_0^{\frac{\pi}{2}} ]

[ = \frac{1}{2} \left( -\frac{1}{2} \cos(\pi) + \frac{1}{2} \cos(0) \right) ]

[ = \frac{1}{2} \left( -\frac{1}{2} \cdot (-1) + \frac{1}{2} \cdot 1 \right) ]

[ = \frac{1}{2} \left( \frac{1}{2} + \frac{1}{2} \right) ]

[ = \frac{1}{4} ]

Therefore, the average value of ( f(x) = \sin(x)\cos(x) ) over the interval ( 0 \leq x \leq \frac{\pi}{2} ) is ( \frac{1}{4} ).

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When evaluating a one-sided limit, you need to be careful when a quantity is approaching zero since its sign is different depending on which way it is approaching zero from. Let us look at some examples.

When evaluating a one-sided limit, you need to be careful when a quantity is approaching zero since its sign is different depending on which way it is approaching zero from. Let us look at some examples.

When evaluating a one-sided limit, you need to be careful when a quantity is approaching zero since its sign is different depending on which way it is approaching zero from. Let us look at some examples.

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