Applications of Integration

Intersections: x² = x + 2 ⇒ x² − x − 2 = 0 ⇒ (x−2)(x+1) = 0, so x = −1 and x = 2.

Which is on top? Test x = 0: line gives 2, parabola gives 0. The line is on top throughout.

Sanity: area is positive, and 4.5 is plausible for a region roughly 3 wide and averaging ~1.5 tall. ✓

Intersections: √x = x/2 ⇒ x = x²/4 ⇒ x² − 4x = 0, so x = 0 and x = 4.

Which is outer? At x = 1: √1 = 1 vs 1/2. The square root is farther from the x-axis, so it is R_out.

The trap: writing π∫(√x − x/2)² dx. That squares the difference instead of differencing the squares, and gives a different (wrong) number.

Rotating about the y-axis with the region described as a function of x ⇒ shells. (Disks would need x in terms of y, which is two-valued here — a genuine mess.)

Integrate by parts with u = x, dv = sin x dx, so v = −cos x:

Note: cos π = −1, so −π(−1) = +π. Sign slips there are the usual cost on this problem.

Square it, being careful with the cross term:

That is the whole trick. The cross term flips sign from −¹⁄₂ to +¹⁄₂ when you add the 1, turning the expression into a perfect square so the radical disappears. Every solvable textbook arc length is built this way.

(a) Hooke: F = kx, so W = ∫₀^dkx dx = ¹⁄₂kd².

(b) Integrate between the two extensions — do not start from zero:

Note the nonlinearity: the second 0.30 m costs 52.5 J while the first 0.20 m cost only 10 J. Work grows as the square of extension, so later stretch is far more expensive. Same reason a capacitor’s last volt costs more energy than its first.

The MVT for integrals guarantees some c in [0,3] with f(c) = 3:

Reject c = −√3 — it is outside the interval. Note c sits right of the midpoint 1.5, because x² is concave up and weights the right end more heavily.

On one period, v(t) = V_pt/T.

Compare: sine gives 0.707V_p, sawtooth 0.577V_p, square 1.000V_p. The ordering reflects how much time each waveform spends near its peak. A square wave is always at full amplitude; a sawtooth spends most of its period well below peak.

Consequence: a 10 V-peak square wave delivers roughly three times the power of a 10 V-peak sawtooth into the same resistor (1.00² vs 0.577², a ratio of 3). Peak voltage alone tells you nothing about delivered power.

|sin| repeats every half period, so average over [0, T/2] where sin ωt ≥ 0:

With ω = 2π/T, we have ωT/2 = π, so the bracket is −cos π + cos 0 = 2:

Why a smoothing capacitor: the average is 0.637V_p, but the instantaneous output still swings all the way from 0 to V_p twice per cycle. That is not DC — it is a pulsating waveform whose average happens to be nonzero. A capacitor stores charge on the peaks and supplies it through the troughs, flattening the ripple toward the peak value. Without it, any circuit downstream sees its supply collapse to zero 120 times a second.

Contrast worth remembering: average = 0.637V_p but RMS = 0.707V_p. Different questions, different integrals — the first asks “what does a DC meter read,” the second asks “how much does it heat a resistor.”

(a) Charge. Charge is the accumulated current:

(b) Voltage. C = 50 mF = 0.050 F:

(c) Energy, method 1:

Method 2, from charge:

Method 3 (the honest one) — integrate power directly, since v(t) = q(t)/C = 0.25t²/0.050 = 5t²:

Three routes, one answer. Unit warning: mF is millifarads (10⁻³), not microfarads. Reading it as µF makes the voltage 80 kV and the energy off by a thousand.

Area first — both centroid formulas divide by it:

Horizontal coordinate:

Vertical coordinate, using ¹⁄₂f²:

Centroid: (2.4, 0.75).

Sanity check both coordinates. The region spans x from 0 to 4 and is fatter on the right, so x̄ = 2.4 should sit right of the midpoint 2. ✓  It spans y from 0 to 2 but is much wider near the bottom, so ȳ = 0.75 should sit well below the midpoint 1. ✓