The length of the arc of \( y = f(x) \) from \( x = a \) to \( x = b \) is
Answer: A
An element of arc satisfies \( ds^{2} = dx^{2} + dy^{2} \), so
\[ ds \;=\; \sqrt{1 + \left(\frac{dy}{dx}\right)^{2}}\,dx, \]
and integrating gives the length. The slope, not the function value or the second derivative, is what enters, since it is the slope that fixes how much longer the curve is than its projection
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The area between \( y = x \) and \( y = x^{2} \) from \( x = 0 \) to \( x = 1 \) is
Answer: D
On this interval the line lies above the parabola, so integrate the difference:
\[ \int_{0}^{1}\left(x - x^{2}\right)dx \;=\; \left[\frac{x^{2}}{2} - \frac{x^{3}}{3}\right]_{0}^{1} \;=\; \frac{1}{2} - \frac{1}{3} \;=\; \frac{1}{6}. \]
The separate areas are \( 1/2 \) and \( 1/3 \), and taking either alone is the usual slip
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The volume swept out by revolving \( y = f(x) \) about the \( x \)-axis between \( a \) and \( b \) is
Answer: C
A thin slice perpendicular to the axis is a disc of radius \( y \) and thickness \( dx \), so its volume is \( \pi y^{2}dx \) and
\[ V \;=\; \pi\int_{a}^{b} y^{2}\,dx. \]
The form \( 2\pi\int y\,dx \) is the surface of revolution's shape rather than the volume's, carrying one power of \( y \) because a surface has one dimension fewer
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Revolving \( y = x \) from \( x = 0 \) to \( x = 3 \) about the \( x \)-axis gives a volume of
Answer: B
Apply the disc formula with \( y = x \):
\[ V \;=\; \pi\int_{0}^{3} x^{2}\,dx \;=\; \pi\left[\frac{x^{3}}{3}\right]_{0}^{3} \;=\; 9\pi. \]
The solid is a cone of radius three and height three, and the elementary formula \( \tfrac{1}{3}\pi r^{2}h = \tfrac{1}{3}\pi(9)(3) = 9\pi \) agrees
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The mean value of \( f \) over \( [a,b] \) is
Answer: A
The mean is the total divided by the length of the interval, so that a constant function returns its own value:
\[ \bar{f} \;=\; \frac{1}{b-a}\int_{a}^{b} f(x)\,dx. \]
Averaging the endpoint values ignores everything between them, and the integral of \( f^{2} \) leads to the root-mean-square instead, which needs a square root as well
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The root-mean-square value of \( \sin \omega t \) over a full cycle is
Answer: D
Square, average over the period and take the root, using \( \sin^{2}\theta = \tfrac{1}{2}(1 - \cos 2\theta) \) whose mean is \( 1/2 \):
\[ \text{r.m.s.} \;=\; \sqrt{\frac{1}{2}} \;=\; \frac{1}{\sqrt{2}} \;\approx\; 0.707. \]
The value \( 2/\pi \approx 0.637 \) is the mean of the *rectified* sine over a half cycle, a different average often confused with this one
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Simpson's one-third rule requires the number of subintervals to be
Answer: C
The rule fits a parabola through each group of three consecutive ordinates, consuming two subintervals at a time, so the total must be even. The three-eighths rule fits a cubic through four ordinates and needs a multiple of three subintervals. Only the trapezoidal rule accepts any number
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The trapezoidal rule with \( n \) strips of width \( h \) estimates the integral as
Answer: B
Each strip contributes the area of a trapezium, \( \tfrac{h}{2}(y_{i} + y_{i+1}) \). Summing, every interior ordinate is counted twice and the two end ordinates once:
\[ I \;\approx\; \frac{h}{2}\left[y_{0} + y_{n} + 2\sum_{i=1}^{n-1} y_{i}\right]. \]
The divisor three and the weight four belong to Simpson's rule
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Estimating \( \displaystyle\int_{0}^{2} x^{2}dx \) by the trapezoidal rule with two strips gives
Answer: A
With \( h = 1 \) the ordinates are \( y_{0} = 0 \), \( y_{1} = 1 \), \( y_{2} = 4 \), so
\[ I \;\approx\; \frac{1}{2}\left[0 + 4 + 2(1)\right] \;=\; 3. \]
The exact value is \( 8/3 \approx 2.667 \), so the rule overestimates by a third. It always does for a convex integrand, the chords lying above the curve
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Estimating that same \( \displaystyle\int_{0}^{2} x^{2}dx \) by Simpson's one-third rule with two strips gives
Answer: D
With the same ordinates and \( h = 1 \),
\[ I \;\approx\; \frac{1}{3}\left[0 + 4(1) + 4\right] \;=\; \frac{8}{3}. \]
Simpson's rule fits a parabola, and the integrand *is* a parabola, so the fit is exact. The rule is in fact exact for any cubic as well, which is why its error term involves the fourth derivative
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