Applications of Integration

An integral adds contributions over an interval. The integrand must represent the quantity per unit of the integration variable. A geometric region’s area is nonnegative even when a graph lies below the horizontal axis.

If a rate R describes change in a quantity, its definite integral over time gives net change. Add the initial amount to obtain the final amount. When inflow and outflow are both given, integrate their difference for net storage change; integrate each separately if the question asks total entering or leaving.

A positive final amount does not imply the amount increased throughout the interval. Extrema of an accumulation function occur among endpoints and places where its net rate is zero or undefined. Use a sign change or compare values to justify a maximum.

The integral of velocity gives displacement. Total distance integrates speed |v|, requiring a split where velocity changes sign if working by hand. Integrating acceleration gives velocity change, not directly position change. Track the level of differentiation carefully.

An integral’s units provide a strong check. Acceleration in ms² integrated over seconds gives ms. If an answer to a distance question still has velocity units, the wrong rate or accumulation was used.

The average value of f on [a,b] is its definite integral divided by interval length. It is not generally the average of endpoint values, except in special cases such as a linear function. A continuous function attains its average value somewhere on the interval.

For a rate, average value has the original rate units. For a temperature function over time, the integral has temperature-times-time units, and dividing by elapsed time returns temperature. The denominator is b − a, not the number of displayed sample points.

With vertical slices, area is the integral of upper minus lower y-values over x. With horizontal slices, use right minus left x-values over y. Find intersections and determine which boundary is larger throughout each subinterval.

A curve can switch roles across an intersection, requiring a split. Squaring a negative difference does not repair an area setup. Choose the slicing direction that gives a clear single-valued description and the fewest necessary pieces.

A solid with known perpendicular cross-sections has volume ∫A(x)dx or the corresponding y integral. If a slice is a square, its area is side²; if it is a semicircle, identify whether the given distance is a diameter or radius before using the formula.

For revolution with washers, subtract inner disk area from outer disk area: π(R² − r²). Radii are distances to the axis, not simply raw function values unless the axis makes that true. Rotation around a shifted horizontal or vertical line requires adjusted distances.

A vertical slice has height top minus bottom and width dx. Find intersections to determine interval boundaries and whether the top curve changes. A horizontal slice instead uses right minus left and dy. Choose the orientation that produces simpler boundaries and fewer pieces, but express every boundary in the chosen variable.

The average value of f on [a,b] is the integral divided by b−a. Its units are the same as f, because the interval width cancels the integration variable’s units. Average value is not generally the average of endpoint values. It is the constant height that would give the same signed area over the interval.

For known cross sections, V=∫A(x)dx or ∫A(y)dy. First determine the length supplied by the base region, then convert that length into area using the stated shape. A square with side s has area s²; an equilateral triangle with side s has area √3 s²4. A semicircle whose diameter is s has radius s2, so its area is πs²8, not πs²2.

For washers, identify the distances from the rotation axis to the outer and inner boundaries. The area is π(R²−r²), not π(R−r)². If the axis is shifted, the radii are distances from that shifted axis. Sketch one representative slice and label both radii before writing the integral; this step prevents most setup errors.

If r(t) is a rate, the final amount equals initial amount plus ∫r(t)dt. The integral alone gives change, not the final amount. When r is velocity, this is displacement; total distance uses ∫|r(t)|dt. Find sign changes before removing the absolute value.

The average value of f on [a,b] is (1(b−a))∫f. It has the same units as f. Average rate of change instead uses endpoint difference divided by interval length. These formulas answer different questions even when both are called an average in ordinary language.

For inflow and outflow, integrate their difference to get net storage change. Integrating only the inflow gives total incoming quantity, not remaining quantity. A maximum stored amount can occur when the net rate changes from positive to negative, requiring a derivative sign argument.

Area between curves uses top minus bottom for vertical slices or right minus left for horizontal slices. Find intersections and determine ordering on each interval. If ordering changes, split the integral or use a correctly handled absolute value.

A sketch helps choose the variable and detect whether one curve changes its role. The bounds are values of the integration variable, not automatically the coordinates first given in the problem. For horizontal slices, rewrite boundaries as x-functions of y where needed.

A numerical answer should have square units. Signed cancellation is not appropriate for geometric area, so a negative area result indicates an ordering or bound problem. Symmetry can simplify work only when the region actually has the claimed symmetry.

Volume is ∫A(x)dx or ∫A(y)dy. Identify the cross-sectional shape and express its area using the slice’s dimensions. For a square, area is side squared; for an equilateral triangle, a different area factor is required. A labeled base length is not necessarily a radius.

For rotation with washers, use π(R²−r²), where radii are distances from the axis of rotation. Rotating around y=3 changes those distances; do not simply square the original curve values. The outer radius is determined geometrically and can switch across intervals.

Check dimensions and limiting cases. A cross-sectional area has square units and integration adds one length unit. If the inner radius is zero, the washer becomes a disk. Sketching one representative slice is often more useful than drawing an elaborate solid.

For a volume with known cross-sections, first draw one cross-section perpendicular to the stated axis. Determine its base or radius from the bounding curves. Then apply the area formula for that shape and integrate along the direction in which the slices accumulate.

If the region between y=x and y=x² on [0,1] forms square cross-sections perpendicular to the x-axis, each side is x−x² and area is (x−x²)². Integrating only x−x² would give the planar region’s area, not the solid’s volume. A dimension check distinguishes the two immediately.

For washers around a horizontal line, radii are vertical distances. Around a vertical line with horizontal slices, radii are horizontal distances. Choose slices that produce a clear area expression and bounds; the picture’s orientation alone does not dictate the variable.

If an amount Q(t) changes at net rate r(t), then Q′=r. Candidate interior extrema occur where r=0 or is undefined, provided Q is defined. A positive-to-negative rate change gives a local maximum of the amount. The largest rate instead indicates the fastest increase, not necessarily the largest amount.

For an absolute maximum over a fixed time interval, compare the amount at all relevant candidates and endpoints. Add the initial value if actual amounts are needed. If only comparisons are required, the same initial constant cancels, but explain that reasoning rather than silently omitting it.

For total distance, total consumption magnitude, or total variation, signs may need absolute values. In a physical inflow–outflow model, distinguish total inflow, total outflow, and net stored change. They can all be positive numbers while representing different integrals.