Source: Integral Calculus, University of Texas at Austin (wolesensky, 54390)
Difficulty: Intermediate to Advanced | Prerequisites: Fundamental Theorem of Calculus (Part 1 and Part 2), chain rule for derivatives, understanding of antiderivatives, trigonometric identities.
This set of topics ties together several ideas from single-variable calculus that appear alongside double integrals in this assignment. Reading contour maps to determine the sign of partial derivatives connects to multivariable thinking. The Fundamental Theorem of Calculus (FTC) problems test whether you can differentiate an integral, including when the upper limit is itself a function. Absolute value integrals require you to split the domain at the point where the integrand changes sign. If you are solid on the FTC and comfortable with piecewise integration, this material is within reach.
Partial derivatives at a point on a contour map are read from the direction of steepest descent. The FTC says the derivative of an integral with a variable upper limit is the integrand evaluated at that limit, with the chain rule applied when the limit is a function. Absolute value integrals require splitting at the zero of the expression inside the bars.
Contour map (level curves)
A plot showing curves where f(x, y) takes constant values. The spacing and direction of these curves tell you how fast and in which direction f is changing. Think of it as a topographic map where each line represents a fixed elevation.
Partial derivative (f_x, f_y)
The rate of change of f with respect to one variable while the other is held constant. f_x measures change in the x-direction, f_y in the y-direction.
Fundamental Theorem of Calculus (FTC), Part 1
If g(x) = integral from a to x of f(t) dt, then g'(x) = f(x). In simple terms, differentiating an integral with respect to its upper limit just gives back the integrand.
FTC with chain rule
If F(x) = (d/dx) integral from a to h(x) of f(t) dt, then F(x) = f(h(x)) * h'(x). The chain rule kicks in because the upper limit is a function of x, not just x itself.
Absolute value integral
An integral of |g(x)|, which requires finding where g(x) = 0, then integrating g(x) on intervals where it is positive and -g(x) where it is negative. Think of it as measuring total distance rather than net displacement.
The contour map shows level curves of f labelled 0, -2, -4, -6. At point P, moving in the positive x-direction takes you from higher to lower contour values (towards more negative values), so f_x < 0. Moving in the positive y-direction, the contour values do not change (P sits on a curve running roughly in the y-direction), so f_y = 0.
Answer: f_x < 0, f_y = 0.
If g(x) is continuous on [a, b] and m is a constant, then the integral of m * g(x) dx from a to b equals m times the integral of g(x) dx from a to b. By FTC, if G'(x) = g(x), then the integral of g(x) dx = G(b) - G(a).
So I = m(G(b) - G(a)), where G'(x) = g(x).
Answer: I = m(G(b) - G(a)), G'(x) = g(x).
Given g(x) = integral from 4 to x of f(t) dt, the FTC gives g'(x) = f(x). The function g has a maximum where g' changes from positive to negative, which means f changes from positive to negative.
From the graph, f is positive from x = 4 to approximately x = 9, then becomes negative. So g is increasing on (4, 9) and decreasing after x = 9. The maximum of g occurs at x = 9.
Answer: x = 9.
Given F(x) = (d/dx) integral from 1 to x^2 of 5t^3 dt.
By the FTC with chain rule: F(x) = 5(x^2)^3 * (d/dx)(x^2) = 5x^6 * 2x = 10x^7.
At x = 1: F(1) = 10(1)^7 = 10.
Answer: F(1) = 10.
The graph shows F, an antiderivative of f, on [0, 8]. Since F' = f, the integral of |f(x)| dx from 0 to 8 measures the total variation of F.
From the graph, F appears to start at about 4, rise to about 8, then fall back to about 4. The total variation is the sum of all the rises and falls: the rise from 4 to 8 is 4, the fall from 8 to 4 is 4. Total = 4 + 4 = 8.
Answer: I = 8.
Evaluate I = 2 * integral from 0 to pi/2 of |sin(theta) - cos(theta)| d(theta).
The expression sin(theta) - cos(theta) = 0 when sin(theta) = cos(theta), which occurs at theta = pi/4.
For 0 <= theta < pi/4: cos(theta) > sin(theta), so |sin - cos| = cos - sin. For pi/4 < theta <= pi/2: sin(theta) > cos(theta), so |sin - cos| = sin - cos.
By symmetry of sine and cosine around pi/4, both pieces give the same value.
First piece: integral from 0 to pi/4 of (cos - sin) d(theta) = [sin + cos] from 0 to pi/4 = (sqrt(2)/2 + sqrt(2)/2) - (0 + 1) = sqrt(2) - 1.
Second piece: integral from pi/4 to pi/2 of (sin - cos) d(theta) = [-cos - sin] from pi/4 to pi/2 = (0 - 1) - (-sqrt(2)/2 - sqrt(2)/2) = -1 + sqrt(2) = sqrt(2) - 1.
Total inside: 2(sqrt(2) - 1). Multiply by 2: I = 4(sqrt(2) - 1).
Answer: I = 4(sqrt(2) - 1).
\frac{d}{dx}\int_a^x f(t)\,dt = f(x)FTC Part 1: the derivative of the integral with respect to its upper limit.
\frac{d}{dx}\int_a^{h(x)} f(t)\,dt = f(h(x))\cdot h'(x)FTC with chain rule: when the upper limit is a function h(x).
\int_a^b |g(x)|\,dx = \int_a^c (-g(x))\,dx + \int_c^b g(x)\,dxAbsolute value integral, where c is the point in (a, b) where g(c) = 0 and g changes sign (negative before c, positive after c in this example).
The FTC with chain rule appears in physics when differentiating accumulated quantities whose limits depend on another variable, for example the work done by a force that depends on a changing displacement. Absolute value integrals compute total distance travelled (as opposed to displacement) and total energy dissipated regardless of direction.
Forgetting the chain rule when the upper limit is x^2 rather than x. The factor h'(x) = 2x must appear.
Confusing where g(x) has a maximum with where g'(x) has a maximum. The maximum of g occurs where g' = f changes sign from positive to negative, not where f itself is largest.
Treating |f(x)| as f(x). The absolute value removes negative contributions, so the integral of |f| is always at least as large as the integral of f.
Reading contour maps backwards: f_x is determined by moving in the x-direction (horizontally), not along the contour line.
The FTC with chain rule (Q16) is a high-frequency exam question. Practise recognising when the upper limit is a function of x.
Absolute value integrals (Q17, Q18) require you to find the splitting point. Always set the expression inside the bars equal to zero first.
Contour map reading (Q13) tests conceptual understanding of partial derivatives. Know that moving in the x-direction tells you about f_x, and moving in the y-direction tells you about f_y.
The constant multiple rule (Q14) seems basic but appears as a trap when students try to integrate m as if it were a function.
True or false: If g(x) = integral from 2 to x of f(t) dt, then g'(x) = f(x). (True, by FTC Part 1.)
Fill in the blank: (d/dx) integral from 1 to x^3 of sin(t) dt = ______. (sin(x^3) * 3x^2.)
True or false: The integral of |f(x)| from a to b can be smaller than the integral of f(x) from a to b. (False. It is always greater than or equal.)
Fill in the blank: On a contour map, f_x > 0 at a point means that moving in the positive x-direction, the contour values ______. (Increase.)
Q: From a contour map where contour values decrease as you move in the positive x-direction at point P, and stay constant in the y-direction, what are the signs of f_x and f_y at P?
A: f_x < 0 (values decrease with x) and f_y = 0 (values unchanged with y).
Q: If F(x) = (d/dx) integral from 1 to x^2 of 5t^3 dt, find F(1).
A: F(x) = 5(x^2)^3 * 2x = 10x^7. At x = 1, F(1) = 10.
Q: Given g(x) = integral from 4 to x of f(t) dt, and f is positive on (4, 9) and negative after 9, where does g have its maximum?
A: At x = 9, where f changes from positive to negative, so g' changes from positive to negative.
Q: Evaluate 2 * integral from 0 to pi/2 of |sin(theta) - cos(theta)| d(theta).
A: Split at theta = pi/4. Each piece gives sqrt(2) - 1. Total = 2 * 2(sqrt(2) - 1) = 4(sqrt(2) - 1).
The FTC with chain rule connects to substitution in integration and to the multivariable chain rule you will use in later chapters. Contour map reading is the visual foundation for gradient vectors and directional derivatives. Absolute value integrals reappear in arc length calculations and in computing total variation of functions.
Fundamental Theorem of Calculus, FTC, FTC Part 1, FTC with chain rule, Leibniz rule, contour map, level curves, partial derivatives, f_x, f_y, absolute value integral, total variation, antiderivative graph, splitting point, piecewise integration, sin minus cos absolute value, integral calculus