The diagram shows part of the curve y=1+sqrt3 sin2x +cos2x. Find the exact value of the area of shaded region. the curve intersect the y axis and x axis for your infomation

Answer 1

If you are looking for the area bounded by the co-ordinate axes and the given curve,

it will be given by the integral from 0 to π/2 of the curve given by y.

If you cannot do the integration, please let me know and I will help with that.

Answer 2

To find the area underneath this curve (between the curve and the x-axis), we need to take a definite integral of the function describing the curve (which you have provided) from the starting x-value to the ending x-value.

The starting x-value is 0, since the y-axis is the left bound of the shaded area. To find the rightmost x-value, we need to set y equal to 0, since that is where the curve crosses the x-axis. This gives us:

0 = 1 + sqrt(3)*sin(2x) +cos(2x)

First, we subtract the 1 over to the left hand side:

-1 = sqrt(3)*sin(2x) +cos(2x)

Next, we are left with the question of how to combine this sine and cosine so we can isolate x. We see that the coefficient on the sine term is a familiar number, sin(pi/3) is sqrt(3)/2. Let's see what things look like if we divide both sides by 2.

-1/2 = sqrt(3)/2*sin(2x) + 1/2*cos(2x)

Now, we can also recognize the coefficient on the cosine term as sin(pi/6) or equivalently cos(pi/3). Using the second form, along with sqrt(3)/2 = sin(pi/3), we can rewrite the right hand side as:

-1/2 = sin(pi/3)*sin(2x) + cos(pi/3)*cos(2x)

Now, we need to thing back to our trigonometric addition/subtraction formulas (http://mathworld.wolfram.com/TrigonometricAdditionFormulas.html). When we have two sines multiplied to each other added to two cosines multiplied to each other, with the same sets of arguments (pi/3 and 2x in this case), we recognize this as a "cosine of differences" (equation 4 on the link provided). We can rewrite our equation as:

-1/2 = cos(pi/3-2x)

Since cosine is an even function, we can rewrite this as

-1/2 = cos(2x-pi/3)

Now, we take the inverse cosine of both sides:

arccos(-1/2) = arccos(cos(2x-pi/3))

arccos(-1/2) = 2x - pi/3

The inverse cosine of -1/2 is equal to 2*pi/3.

2*pi/3 = 2x - pi/3

Isolating x:

x = pi/2

All of this was to tell us that our integral bounds are from 0 to pi/2.

The integral of our equation above is:

x - sqrt(3)/2*cos(2x) + sin(2x)

At pi/2, this equals

pi/2 - sqrt(3)/2*cos(pi) + sin(pi) = pi/2 + sqrt(3)/2

At 0, this equals

0 - sqrt(3)/2*cos(0) + sin(0) = - sqrt(3)/2

Subtracting the first result from the second gives us the definite integral and the area under the curve

A = pi/2 + sqrt(3)/2 - (-sqrt(3)/2)

A = pi/2 + sqrt(3)