The diagram shows part of the curve y=2sin⁡2x+1y = 2 \sin 2x + 1. PP is the maximum point of the curve and QQ is the point on the curve at which the gradient of the tangent is −4-4.

  1. (a)

    Find the coordinates of PP and of QQ.

  2. (b)

    Find the area of the shaded region bounded by the curve, the axes, line PQPQ and vertical line QRQR. Leave your answer in the exact form.

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Worked solution

10(a)

1. Find the coordinates of the maximum point $P$ using the properties of the sine function.

For y=2sin⁡2x+1y = 2 \sin 2x + 1, the maximum value occurs when sin⁡2x=1\sin 2x = 1.
ymax⁡=2(1)+1=3y_{\max} = 2(1) + 1 = 3
sin⁡2x=1  ⟹  2x=π2  ⟹  x=π4\sin 2x = 1 \implies 2x = \frac{\pi}{2} \implies x = \frac{\pi}{4}
Coordinates of PP are (π4,3)\left(\frac{\pi}{4}, 3\right).

2. Differentiate the equation of the curve to find the gradient function.

dydx=2(2cos⁡2x)=4cos⁡2x\frac{dy}{dx} = 2(2 \cos 2x) = 4 \cos 2x

3. Equate the gradient function to $-4$ to find the $x$-coordinate of $Q$.

At QQ, dydx=−4\frac{dy}{dx} = -4
4cos⁡2x=−44 \cos 2x = -4
cos⁡2x=−1\cos 2x = -1
2x=π  ⟹  x=π22x = \pi \implies x = \frac{\pi}{2}

4. Find the corresponding $y$-coordinate of $Q$.

When x=π2x = \frac{\pi}{2},
y=2sin⁡(2×π2)+1=2sin⁡π+1=2(0)+1=1y = 2 \sin\left(2 \times \frac{\pi}{2}\right) + 1 = 2 \sin \pi + 1 = 2(0) + 1 = 1
Coordinates of QQ are (π2,1)\left(\frac{\pi}{2}, 1\right).

Answer

P(π4,3)P\left(\frac{\pi}{4}, 3\right), Q(π2,1)Q\left(\frac{\pi}{2}, 1\right)

10(b)

1. State the coordinates of $P$ and $Q$ from part (a)

From part (a),
P=(π4,3)P = \left(\frac{\pi}{4}, 3\right)
Q=(π2,1)Q = \left(\frac{\pi}{2}, 1\right)

2. Set up the expression for the total shaded area

The shaded area is the sum of the area under the curve from x=0x = 0 to x=π4x = \frac{\pi}{4} and the area of the trapezium under the line PQPQ from x=π4x = \frac{\pi}{4} to x=π2x = \frac{\pi}{2}.
Area=∫0π4(2sin⁡2x+1) dx+Area of trapezium\text{Area} = \int_{0}^{\frac{\pi}{4}} (2 \sin 2x + 1) \, dx + \text{Area of trapezium}

3. Evaluate the definite integral for the area under the curve

∫0π4(2sin⁡2x+1) dx=[−cos⁡2x+x]0π4\int_{0}^{\frac{\pi}{4}} (2 \sin 2x + 1) \, dx = \left[ -\cos 2x + x \right]_{0}^{\frac{\pi}{4}}
=(−cos⁡(π2)+π4)−(−cos⁡(0)+0)= \left( -\cos\left(\frac{\pi}{2}\right) + \frac{\pi}{4} \right) - (-\cos(0) + 0)
=(0+π4)−(−1)= \left( 0 + \frac{\pi}{4} \right) - (-1)
=1+π4= 1 + \frac{\pi}{4}

4. Calculate the area of the trapezium and find the total area

Area of trapezium=12×(3+1)×(π2−π4)\text{Area of trapezium} = \frac{1}{2} \times (3 + 1) \times \left(\frac{\pi}{2} - \frac{\pi}{4}\right)
=12×4×π4=π2= \frac{1}{2} \times 4 \times \frac{\pi}{4} = \frac{\pi}{2}
Total Area=1+π4+π2=1+3π4\text{Total Area} = 1 + \frac{\pi}{4} + \frac{\pi}{2} = 1 + \frac{3\pi}{4}

Answer

1+3π41 + \frac{3\pi}{4}

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