Question Statement

Find the area bounded by the graph:

(i) $y=1+\cos x$ on the interval $[0,3\pi ]$

(ii) $y=-1+\sin x$ on the interval $\left[-\frac{3\pi }{2},\frac{3\pi }{2}\right]$

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Background and Explanation

The area bounded by a curve $y=f(x)$ and the x-axis over an interval $[a,b]$ is given by the definite integral ${\int }_a^{b}f(x)dx$ when the curve lies above the x-axis. If the curve lies below the x-axis, we must take the absolute value (negate the integral) to obtain a positive area measurement.

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Solution

Part (i): $y=1+\cos x$ on $[0,3\pi ]$

Since $\cos x\ge -1$ for all $x$, we have $y=1+\cos x\ge 0$. The graph lies entirely above the x-axis, so the area is:

$\begin{array}{rl}\text{Area}& ={\int }_0^{3\pi }ydx\\ & ={\int }_0^{3\pi }(1+\cos x)dx\\ & ={\int }_0^{3\pi }1dx+{\int }_0^{3\pi }\cos xdx\\ & ={\left[x\right]}_0^{3\pi }+{\left[\sin x\right]}_0^{3\pi }\\ & =(3\pi -0)+(\sin 3\pi -\sin 0)\\ & =3\pi +(0-0)\\ & =3\pi \end{array}$

Part (ii): $y=-1+\sin x$ on $\left[-\frac{3\pi }{2},\frac{3\pi }{2}\right]$

Since $-1\le \sin x\le 1$, we have $-2\le -1+\sin x\le 0$. The graph lies entirely below (or on) the x-axis. Therefore, we take the absolute value by negating the integral:

$\begin{array}{rl}\text{Area}& ={\int }_{-3\pi /2}^{3\pi /2}|y|dx\\ & =-{\int }_{-3\pi /2}^{3\pi /2}(-1+\sin x)dx\\ & ={\int }_{-3\pi /2}^{3\pi /2}1dx-{\int }_{-3\pi /2}^{3\pi /2}\sin xdx\\ & ={\left[x\right]}_{-3\pi /2}^{3\pi /2}+{\left[(-\cos x)\right]}_{-3\pi /2}^{3\pi /2}\\ & =\left(\frac{3\pi }{2}-\left(-\frac{3\pi }{2}\right)\right)-\left(\cos \frac{3\pi }{2}-\cos \left(-\frac{3\pi }{2}\right)\right)\\ & =\frac{3\pi }{2}+\frac{3\pi }{2}-(0-0)\\ & =3\pi \end{array}$

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Key Formulas or Methods Used

• Definite integral for area: $\text{Area}={\int }_a^{b}f(x)dx$ (when $f(x)\ge 0$)
• Absolute value adjustment: $\text{Area}=-{\int }_a^{b}f(x)dx$ (when $f(x)\le 0$ on $[a,b]$)
• Antiderivatives: $\int 1dx=x$, $\int \cos xdx=\sin x$, $\int \sin xdx=-\cos x$
• Fundamental Theorem of Calculus: ${\int }_a^{b}f(x)dx=F(b)-F(a)$ where $F$ is the antiderivative of $f$
• Trigonometric values: $\sin (3\pi )=\sin (0)=0$ and $\cos (\pm 3\pi /2)=0$

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Summary of Steps

1. Identify the interval and determine if the function is positive or negative throughout
2. Set up the definite integral, adding a negative sign if the graph lies below the x-axis
3. Split the integral into simpler terms (constant and trigonometric parts)
4. Integrate each term using standard antiderivative rules
5. Evaluate the bounds by substituting upper and lower limits
6. Simplify using known trigonometric values at multiples of $\pi /2$
7. Combine terms to obtain the final area (both parts equal $3\pi$ square units)