The curve CC has equation y=xcos⁡2xy = x\cos 2x, where 0≤x≤π0 \le x \le \pi.

  1. (i)

    Find the exact xx-coordinates of the points where CC crosses the xx-axis.

  2. (ii)

    Sketch CC, stating the coordinates of any points where the curve crosses the xx- and yy-axes.

  3. (iii)

    Find the exact value of
    ∫π4π∣xcos⁡2x∣ dx.\displaystyle \int_{\tfrac{\pi}{4}}^{\pi} |x\cos 2x|\,dx.

  4. (iv)

    Find the volume of revolution when the region bounded by the curve CC, the xx-axis and the line x=πx = \pi is rotated completely about the xx-axis.

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Worked solution

11(i)

1. Set $y = 0$ to find the points of intersection with the $x$-axis.

xcos⁡2x=0x \cos 2x = 0

2. Solve for the first factor, $x = 0$.

x=0x = 0

3. Solve for the second factor, $\cos 2x = 0$, within the given domain $0 \le x \le \pi$.

For 0≤x≤π0 \le x \le \pi, the range for 2x2x is 0≤2x≤2π0 \le 2x \le 2\pi.
cos⁡2x=0\cos 2x = 0
2x=π2,3π22x = \frac{\pi}{2}, \frac{3\pi}{2}
x=π4,3π4x = \frac{\pi}{4}, \frac{3\pi}{4}

Answer

x=0,π4,3π4x = 0, \frac{\pi}{4}, \frac{3\pi}{4}

11(ii)

1. State the coordinates of the points where the curve crosses the axes.

From part (i), setting y=0y = 0 for 0≤x≤π0 \le x \le \pi yields x=0,π4,3π4x = 0, \frac{\pi}{4}, \frac{3\pi}{4}.
When x=0x = 0, y=0y = 0.
Thus, the curve meets or crosses the axes at (0,0)(0, 0), (π4,0)\left(\frac{\pi}{4}, 0\right), and (3π4,0)\left(\frac{3\pi}{4}, 0\right).

2. Determine the behavior and endpoints to complete the sketch.

The endpoint is at x=πx = \pi, where y=πcos⁡(2π)=πy = \pi \cos(2\pi) = \pi.
Using a graphing calculator to observe the shape over 0≤x≤π0 \le x \le \pi, the curve starts at (0,0)(0,0), increases to a local maximum, decreases to cross the xx-axis at π4\frac{\pi}{4}, reaches a local minimum, increases to cross the xx-axis at 3π4\frac{3\pi}{4}, and terminates at the point (π,π)(\pi, \pi).

Answer

Coordinates: (0,0)(0, 0), (π4,0)\left(\frac{\pi}{4}, 0\right), (3π4,0)\left(\frac{3\pi}{4}, 0\right)

[A sketch of a smooth curve starting at (0,0)(0,0), reaching a local maximum, crossing the xx-axis at π4\frac{\pi}{4}, reaching a local minimum, crossing the xx-axis at 3π4\frac{3\pi}{4}, and ending at (π,π)(\pi, \pi).]

11(iii)

1. Identify the interval where the integrand is negative and split the definite integral to resolve the modulus.

From part (i), xcos⁡2x=0x\cos 2x = 0 at x=π4x = \frac{\pi}{4} and x=3π4x = \frac{3\pi}{4}.
For x∈[π4,3π4]x \in \left[\frac{\pi}{4}, \frac{3\pi}{4}\right], cos⁡2x≤0\cos 2x \le 0, so xcos⁡2x≤0x\cos 2x \le 0.
For x∈[3π4,π]x \in \left[\frac{3\pi}{4}, \pi\right], cos⁡2x≥0\cos 2x \ge 0, so xcos⁡2x≥0x\cos 2x \ge 0.
∫π4π∣xcos⁡2x∣ dx=∫π43π4−xcos⁡2x dx+∫3π4πxcos⁡2x dx\int_{\frac{\pi}{4}}^{\pi} |x\cos 2x|\,dx = \int_{\frac{\pi}{4}}^{\frac{3\pi}{4}} -x\cos 2x \,dx + \int_{\frac{3\pi}{4}}^{\pi} x\cos 2x \,dx

2. Perform integration by parts to find the indefinite integral of $x\cos 2x$.

Let u=x  ⟹  dudx=1u = x \implies \frac{du}{dx} = 1.
Let dvdx=cos⁡2x  ⟹  v=12sin⁡2x\frac{dv}{dx} = \cos 2x \implies v = \frac{1}{2}\sin 2x.
∫xcos⁡2x dx=x(12sin⁡2x)−∫12sin⁡2x dx\int x\cos 2x \,dx = x\left(\frac{1}{2}\sin 2x\right) - \int \frac{1}{2}\sin 2x \,dx
=12xsin⁡2x−(−14cos⁡2x)=12xsin⁡2x+14cos⁡2x= \frac{1}{2}x\sin 2x - \left( -\frac{1}{4}\cos 2x \right) = \frac{1}{2}x\sin 2x + \frac{1}{4}\cos 2x

3. Evaluate the first definite integral over the interval $[ rac{\pi}{4}, \frac{3\pi}{4}]$.

∫π43π4−xcos⁡2x dx=−[12xsin⁡2x+14cos⁡2x]π43π4\int_{\frac{\pi}{4}}^{\frac{3\pi}{4}} -x\cos 2x \,dx = -\left[ \frac{1}{2}x\sin 2x + \frac{1}{4}\cos 2x \right]_{\frac{\pi}{4}}^{\frac{3\pi}{4}}
=−[(12(3π4)sin⁡(3π2)+14cos⁡(3π2))−(12(π4)sin⁡(π2)+14cos⁡(π2))]= -\left[ \left( \frac{1}{2}\left(\frac{3\pi}{4}\right)\sin\left(\frac{3\pi}{2}\right) + \frac{1}{4}\cos\left(\frac{3\pi}{2}\right) \right) - \left( \frac{1}{2}\left(\frac{\pi}{4}\right)\sin\left(\frac{\pi}{2}\right) + \frac{1}{4}\cos\left(\frac{\pi}{2}\right) \right) \right]
=−[(3π8(−1)+0)−(π8(1)+0)]= -\left[ \left( \frac{3\pi}{8}(-1) + 0 \right) - \left( \frac{\pi}{8}(1) + 0 \right) \right]
=−(−3π8−π8)=4π8=π2= -\left( -\frac{3\pi}{8} - \frac{\pi}{8} \right) = \frac{4\pi}{8} = \frac{\pi}{2}

4. Evaluate the second definite integral over the interval $[ rac{3\pi}{4}, \pi]$ and sum the results.

∫3π4πxcos⁡2x dx=[12xsin⁡2x+14cos⁡2x]3π4π\int_{\frac{3\pi}{4}}^{\pi} x\cos 2x \,dx = \left[ \frac{1}{2}x\sin 2x + \frac{1}{4}\cos 2x \right]_{\frac{3\pi}{4}}^{\pi}
=(12(π)sin⁡(2π)+14cos⁡(2π))−(12(3π4)sin⁡(3π2)+14cos⁡(3π2))= \left( \frac{1}{2}(\pi)\sin(2\pi) + \frac{1}{4}\cos(2\pi) \right) - \left( \frac{1}{2}\left(\frac{3\pi}{4}\right)\sin\left(\frac{3\pi}{2}\right) + \frac{1}{4}\cos\left(\frac{3\pi}{2}\right) \right)
=(0+14(1))−(−3π8)=14+3π8= \left( 0 + \frac{1}{4}(1) \right) - \left( -\frac{3\pi}{8} \right) = \frac{1}{4} + \frac{3\pi}{8}
Total integral value:
π2+(14+3π8)=4π8+3π8+14=7π8+14\frac{\pi}{2} + \left( \frac{1}{4} + \frac{3\pi}{8} \right) = \frac{4\pi}{8} + \frac{3\pi}{8} + \frac{1}{4} = \frac{7\pi}{8} + \frac{1}{4}

Answer

7π8+14\frac{7\pi}{8} + \frac{1}{4}

11(iv)

1. Identify the correct region of integration and set up the volume integral.

From part (i), the xx-intercepts of CC are x=0,π4x = 0, \frac{\pi}{4}, and 3π4\frac{3\pi}{4}.
The specific region bounded by the curve CC, the xx-axis, and the vertical line x=πx = \pi lies between the final intercept and x=πx = \pi, which is the interval [3π4,π]\left[\frac{3\pi}{4}, \pi\right].
Setting up the volume of revolution about the xx-axis:
V=π∫3π4πy2 dx=π∫3π4π(xcos⁡2x)2 dxV = \pi \int_{\frac{3\pi}{4}}^{\pi} y^2 \,dx = \pi \int_{\frac{3\pi}{4}}^{\pi} (x\cos 2x)^2 \,dx

2. Evaluate the definite integral using a graphing calculator.

Using a graphing calculator to evaluate the integral numerically:
V≈10.465V \approx 10.465
V=10.5 (to 3 s.f.)V = 10.5 \text{ (to 3 s.f.)}

Answer

10.510.5

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