# What is the net area between #f(x)=sinxcosx# in #x in[0,2pi] # and the x-axis?

We may rewrite this function by making use of the double angle trig identities :

We now draw the graph of this function to help decide limits of integration

graph{1/2sin(2x) [-4.933, 4.934, -2.466, 2.467]}

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To find the net area between ( f(x) = \sin(x)\cos(x) ) and the x-axis on the interval ([0, 2\pi]), you need to evaluate the definite integral of the absolute value of ( f(x) ) over that interval.

[ \text{Net Area} = \int_{0}^{2\pi} |f(x)| , dx ]

[ = \int_{0}^{2\pi} |\sin(x)\cos(x)| , dx ]

[ = \int_{0}^{\pi} \sin(x)\cos(x) , dx + \int_{\pi}^{2\pi} -\sin(x)\cos(x) , dx ]

[ = \int_{0}^{\pi} \frac{1}{2}\sin(2x) , dx - \int_{\pi}^{2\pi} \frac{1}{2}\sin(2x) , dx ]

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

[ = \frac{1}{4} \left(1 - (-1) - (-1) - 1\right) ]

[ = \frac{1}{4} \times 2 ]

[ = \frac{1}{2} ]

So, the net area between ( f(x) = \sin(x)\cos(x) ) and the x-axis on the interval ([0, 2\pi]) is ( \frac{1}{2} ).

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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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