How do you find the area lying above the x axis of #y=sinxcosx# for #-pi<=x<=pi#?
for these, see the graph below which gives:
which because of symmetry can be aggregated as
graph{sinx cosx [-3.897, 3.898, -1.95, 1.947]}
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To find the area above the x-axis of (y = \sin(x)\cos(x)) for (-\pi \leq x \leq \pi), follow these steps:
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Simplify the Function: The function (y = \sin(x)\cos(x)) can be simplified using the double-angle formula: (\sin(2x) = 2\sin(x)\cos(x)). Therefore, (y = \sin(x)\cos(x)) can be rewritten as (y = \frac{1}{2}\sin(2x)).
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Determine the Intervals where (y) is Above the x-axis: The function (\sin(2x)) is positive in the intervals where its angle, (2x), lies in the first and second quadrants of the unit circle. For (-\pi \leq x \leq \pi), this corresponds to two intervals for (x): (-\frac{\pi}{2} < x < 0) and (0 < x < \frac{\pi}{2}).
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Integrate the Function Over These Intervals: The area above the x-axis can be found by integrating (y = \frac{1}{2}\sin(2x)) over the intervals where (y) is positive.
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For (0 < x < \frac{\pi}{2}): [ A_1 = \int_{0}^{\frac{\pi}{2}} \frac{1}{2}\sin(2x) , dx ]
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For (-\frac{\pi}{2} < x < 0), (y) is also positive. However, due to the symmetry of the function (y = \frac{1}{2}\sin(2x)) about the y-axis, the area in this interval is the same as the area for (0 < x < \frac{\pi}{2}). Thus, you only need to calculate (A_1) and then multiply by 2 to account for both intervals.
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Calculate the Integral: [ A_1 = \int_{0}^{\frac{\pi}{2}} \frac{1}{2}\sin(2x) , dx = \left. -\frac{1}{4}\cos(2x) \right|_{0}^{\frac{\pi}{2}} = \left(-\frac{1}{4}\cos(\pi) + \frac{1}{4}\cos(0)\right) = \frac{1}{2} ]
Since the area is the same for both positive intervals due to symmetry, the total area above the x-axis is (2 \times A_1 = 2 \times \frac{1}{2} = 1).
Thus, the area lying above the x-axis of (y = \sin(x)\cos(x)) for (-\pi \leq x \leq \pi) is 1 square unit.
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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.
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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