The Workbook

Use \(\int x^n\,dx = \dfrac{x^{n+1}}{n + 1} + C\) for \(n \ne -1\).

1. \(\dfrac{x^4}{4} + C\).

2. \(\ln x + C\). The power rule fails for \(n = -1\); the logarithm takes its place.

3. \(\sin x + C\), because \(\dfrac{d}{dx}\sin x = \cos x\).

4. \(\dfrac{e^{2x}}{2} + C\). Check: differentiating brings down the \(2\).

1. \(x^3 + x^2 + C\).

2. \(\int x^{1/2}\,dx = \dfrac{x^{3/2}}{3/2} + C = \dfrac{2}{3}x^{3/2} + C\).

3. \(\int x^{-2}\,dx = \dfrac{x^{-1}}{-1} + C = -\dfrac{1}{x} + C\).

4. \(-\dfrac{\cos 2t}{2} + C\). Check: \(\dfrac{d}{dt}\left(-\dfrac{\cos 2t}{2}\right) = \sin 2t\).

1. \(\left[\dfrac{x^3}{3}\right]_0^2 = \dfrac{8}{3}\).

2. \(\left[2\sqrt{x}\right]_1^4 = 4 - 2 = 2\).

3. \(\left[\sin x\right]_0^{\pi/2} = 1 - 0 = 1\).

\(\int_0^{\pi}\sin x\,dx = [-\cos x]_0^{\pi} = -(-1) - (-1) = 2\).

\(\int_0^{2\pi}\sin x\,dx = [-\cos x]_0^{2\pi} = -1 + 1 = 0\).

From \(0\) to \(\pi\) the curve is above the axis and the area counts \(+2\). From \(\pi\) to \(2\pi\) the curve is below the axis and the area counts \(-2\). A definite integral adds signed areas, so the total is zero.

1. Area \(= \int_0^3 x^2\,dx\).

2. \(\left[\dfrac{x^3}{3}\right]_0^3 = 9\).

3. The chord from \((0, 0)\) to \((3, 9)\) lies above the curve, because the curve bends upward. The region under the curve is inside the triangle, so its area is less than \(13.5\).

\(\int_0^4 2x\,dx = [x^2]_0^4 = 16\).

The region is a triangle of base \(4\) and height \(2 \times 4 = 8\): area \(= \dfrac{1}{2} \times 4 \times 8 = 16\). The two agree.

Between \(0\) and \(1\) the line \(y = x\) lies above \(y = x^2\). Area \(= \int_0^1 (x - x^2)\,dx = \left[\dfrac{x^2}{2} - \dfrac{x^3}{3}\right]_0^1 = \dfrac{1}{2} - \dfrac{1}{3} = \dfrac{1}{6}\).

With \(u = 1 + x^2\), \(du = 2x\,dx\). When \(x = 0\), \(u = 1\); when \(x = 1\), \(u = 2\). The integral becomes \(\int_1^2 u\,du = \left[\dfrac{u^2}{2}\right]_1^2 = 2 - \dfrac{1}{2} = \dfrac{3}{2}\).

Check: \(\int_0^1 (2x + 2x^3)\,dx = \left[x^2 + \dfrac{x^4}{2}\right]_0^1 = 1 + \dfrac{1}{2} = \dfrac{3}{2}\).

1. \(\Delta x = \int_0^2 3t^2\,dt = [t^3]_0^2 = 8\) m.

2. \(\bar v = \dfrac{\Delta x}{\Delta t} = \dfrac{8}{2} = 4\) m/s.

1. \(v = 0\) when \(10 - 2t = 0\), at \(t = 5\) s.

2. \(\int_0^5 (10 - 2t)\,dt = [10t - t^2]_0^5 = 50 - 25 = 25\) m.

3. \(\int_0^8 (10 - 2t)\,dt = 80 - 64 = 16\) m. This is the displacement.

   From \(5\) s to \(8\) s the velocity is negative and the particle moves back: \(\int_5^8 (10 - 2t)\,dt = (80 - 64) - (50 - 25) = -9\) m. The distance travelled is \(25 + 9 = 34\) m.

1. Area of the trapezium: triangle \(\dfrac{1}{2} \times 2 \times 10 = 10\) m, rectangle \(4 \times 10 = 40\) m, triangle \(\dfrac{1}{2} \times 2 \times 10 = 10\) m. Total \(60\) m.

2. Slope of the first segment: \(a = \dfrac{10 - 0}{2} = 5\) m/s\(^2\).

3. \(\bar v = \dfrac{60}{8} = 7.5\) m/s.

1. Rectangle: \(q = 2 \times 3 = 6\) C.

2. Triangle: \(q = \dfrac{1}{2} \times 2 \times 2 = 2\) C.

3. \(6 + 2 = 8\) C.

1. \(v = 2 + \int_0^t 6t\,dt = 2 + 3t^2\). At \(t = 2\) s: \(v = 2 + 12 = 14\) m/s.

2. \(x = \int_0^t (2 + 3t^2)\,dt = 2t + t^3\). At \(t = 2\) s: \(x = 4 + 8 = 12\) m.

1. \(q = \int_0^1 5\sin\pi t\,dt = \left[-\dfrac{5\cos\pi t}{\pi}\right]_0^1 = \dfrac{5}{\pi}(1 + 1) = \dfrac{10}{\pi} \approx 3.2\) C.

2. \(\int_0^2 5\sin\pi t\,dt = \left[-\dfrac{5\cos\pi t}{\pi}\right]_0^2 = 0\). During the second second the current is negative, so the same charge flows back. The net charge is zero.

1. Triangle: \(W = \dfrac{1}{2} \times 4 \times 20 = 40\) J.

2. Rectangle: \(W = 20 \times 2 = 40\) J.

3. \(40 + 40 = 80\) J.

1. \(F = kx\), so \(k\) is the slope: \(k = \dfrac{40}{0.2} = 200\) N/m.

2. \(W = \int_0^x kx\,dx = k\left[\dfrac{x^2}{2}\right]_0^x = \dfrac{1}{2}kx^2\). This is the area of the triangle under the line.

3. \(W = \dfrac{1}{2} \times 200 \times 0.04 = 4\) J. Check: \(\dfrac{1}{2} \times 0.2 \times 40 = 4\) J.

\(W = \int_0^2 (2x + 3)\,dx = [x^2 + 3x]_0^2 = 4 + 6 = 10\) J.

\(W = \int_1^3 \dfrac{6}{x^2}\,dx = \left[-\dfrac{6}{x}\right]_1^3 = -2 + 6 = 4\) J.

The applied force must balance gravity at each \(r\). \(W = \int_R^{\infty}\dfrac{GMm}{r^2}\,dr = GMm\left[-\dfrac{1}{r}\right]_R^{\infty} = GMm\left(0 + \dfrac{1}{R}\right) = \dfrac{GMm}{R}\).

The integral converges because the force falls as \(1/r^2\); a finite amount of work removes the body completely.

A piece of length \(dx\) at \(x\) has mass \(\lambda\,dx = 3x\,dx\). \(M = \int_0^2 3x\,dx = \left[\dfrac{3x^2}{2}\right]_0^2 = 6\) kg.

\(E = \int_0^4 P\,dt = \int_0^4 20t\,dt = [10t^2]_0^4 = 160\) J.

1. \(\bar v = \dfrac{1}{2}\int_0^2 3t^2\,dt = \dfrac{8}{2} = 4\) m/s, the same as the average velocity in Question 9. The average of \(v\) over time is the displacement divided by the time.

2. \(\dfrac{1}{\pi}\int_0^{\pi}\sin\theta\,d\theta = \dfrac{1}{\pi}[-\cos\theta]_0^{\pi} = \dfrac{1}{\pi}(1 + 1) = \dfrac{2}{\pi}\).

\(A = \int_0^R 2\pi r\,dr = 2\pi\left[\dfrac{r^2}{2}\right]_0^R = \pi R^2\).

Each ring is a thin strip of length \(2\pi r\) and width \(dr\). Adding all the strips from the centre to the rim gives the disc.

\(\int_0^T\sin^2\omega t\,dt = \int_0^T \dfrac{1 - \cos 2\omega t}{2}\,dt = \dfrac{T}{2} - \left[\dfrac{\sin 2\omega t}{4\omega}\right]_0^T\).

With \(\omega T = 2\pi\), \(\sin 2\omega T = \sin 4\pi = 0\). So the integral is \(\dfrac{T}{2}\). The average of \(\sin^2\) over a full cycle is \(\dfrac{1}{2}\).