How do you find the area lying above the x axis of #y=sinxcosx# for #-pi<=x<=pi#?

Answer 1

#1#

you are looking for the integral of #y = sin x cos x (= 1/2 sin 2 x)# within certain limits as we want only the area above the x axis #x in [-pi, pi]#.

for these, see the graph below which gives:

#A = 1/2 int_(-pi)^(-pi/2) sin 2x + 1/2 int_(0)^(pi/2) sin 2x#

which because of symmetry can be aggregated as

#A = int_(0)^(pi/2) sin 2x#
#= [ - 1/2 cos 2x ]_(0)^(pi/2) #
#= 1/2 [ cos 2x ]_(pi/2)^(0) #
#= 1/2 ( cos 0 - cos pi ) = 1#

graph{sinx cosx [-3.897, 3.898, -1.95, 1.947]}

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Answer 2

To find the area above the x-axis of (y = \sin(x)\cos(x)) for (-\pi \leq x \leq \pi), follow these steps:

  1. 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)).

  2. 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}).

  3. 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.

    • For (0 < x < \frac{\pi}{2}): [ A_1 = \int_{0}^{\frac{\pi}{2}} \frac{1}{2}\sin(2x) , dx ]

    • 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.

  4. 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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Answer from HIX Tutor

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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