Difficulty: Intermediate to Advanced | Prerequisites: FTC Parts 1 and 2, basic derivative rules (including chain rule), definite integral properties.
The substitution rule (u-substitution) is the integration counterpart of the chain rule for derivatives. It is the first systematic technique for finding antiderivatives of composite functions. Once you can handle substitution, you can evaluate a far wider range of integrals than the basic power/trig rules allow. Area between curves then applies definite integrals to a new geometric question: how much space sits between two graphs? Together, these topics round out the core toolkit of Calculus I integration.
U-substitution lets you undo the chain rule inside an integral: set u equal to the inner function, replace dx with du, and simplify. For definite integrals you can also change the limits to match u. Even/odd symmetry shortcuts save time on symmetric intervals. Area between two curves equals the integral of the top function minus the bottom (or right minus left in the y-direction).
Substitution rule (u-substitution)
An integration technique that reverses the chain rule: ∫ f(g(x)) · g'(x) dx = ∫ f(u) du, where u = g(x). Think of it as spotting a function and its derivative sitting together inside the integral, then swapping them for a single variable u.
Change of limits (for definite integrals)
When applying u-substitution to a definite integral, convert the x-limits to u-limits: if u = g(x), then the new lower limit is g(a) and the new upper limit is g(b). This avoids having to substitute back to x.
Even function
A function satisfying f(–x) = f(x) for all x. Its graph is symmetric about the y-axis. Examples: x², cos(x), |x|.
Odd function
A function satisfying f(–x) = –f(x) for all x. Its graph has rotational symmetry about the origin. Examples: x³, sin(x), tan(x).
Area between curves
The total region enclosed between two continuous functions f(x) and g(x) from x = a to x = b. Computed as ∫ (from a to b) |f(x) – g(x)| dx. In simple terms, integrate "top minus bottom" (or "right minus left" in the y-direction).
Integrating with respect to y
When two curves are more naturally expressed as x = f(y) and x = g(y), it can be simpler to integrate along the y-axis. The area is ∫ (from c to d) |f(y) – g(y)| dy, where c and d are the y-bounds.
The theorem: if u = g(x) is differentiable and f is continuous on the range of g, then
∫ f(g(x)) · g'(x) dx = ∫ f(u) du
Step-by-step method:
Identify the "inner function" and call it u.
Compute du = g'(x) dx.
Rewrite the entire integral in terms of u and du. Every x must vanish.
Integrate with respect to u.
Substitute back: replace u with g(x).
Look for a composition f(g(x)) where g'(x) (or a constant multiple of it) is also present as a factor. Common patterns:
∫ x² √(x³ + 1) dx: set u = x³ + 1, du = 3x² dx
∫ dt / (1 – 3t)⁵: set u = 1 – 3t, du = −3 dt
∫ sin(2x + 1) dx: set u = 2x + 1, du = 2 dx
∫ cos²(x) sin(x) dx: set u = cos(x), du = –sin(x) dx
If your choice of u leaves an x behind that you cannot express in terms of u, try a different u or solve u = g(x) for x and substitute.
Approach A, change limits: replace the x-bounds with u-bounds.
∫ (from a to b) f(g(x)) g'(x) dx = ∫ (from g(a) to g(b)) f(u) du
You never need to substitute back. This is cleaner and is the method exams often require.
Approach B, substitute back: evaluate the indefinite integral in u, convert back to x, then plug in the original x-limits.
Both give the same answer. Use whichever the problem specifies.
Sometimes after setting u = g(x) you need to solve for x in terms of u to replace leftover x factors. For example, ∫ x² √(x + 2) dx with u = x + 2 gives x = u – 2 and x² = (u – 2)². Expand, multiply through, and integrate term by term.
f is even if f(–x) = f(x). Graph is symmetric about the y-axis.
f is odd if f(–x) = –f(x). Graph is symmetric about the origin.
If f is continuous on [–a, a]:
f even: ∫ (from –a to a) f(x) dx = 2 ∫ (from 0 to a) f(x) dx
f odd: ∫ (from –a to a) f(x) dx = 0
The odd-function rule is particularly useful. If you spot that the integrand is odd and the limits are symmetric, the answer is zero immediately, no calculation needed.
When an integrand on [–a, a] contains both even and odd pieces, split by linearity:
∫ (from –π/4 to π/4) (1 + x + x² tan x) dx
Here 1 is even, x² tan(x) is odd (product of even and odd), and x is odd. The odd parts integrate to zero, leaving 2 ∫ (from 0 to π/4) 1 dx = π/2.
If f(x) ≥ g(x) on [a, b], the area between the two curves is:
Area = ∫ (from a to b) [f(x) – g(x)] dx
When the curves cross (so neither is always on top), use:
Area = ∫ (from a to b) |f(x) – g(x)| dx
In practice, find the crossing points (set f = g and solve), split the interval at those points, determine which function is on top in each piece, and integrate each piece with the correct order.
If the problem says "find the area enclosed by y = f(x) and y = g(x)", the bounds are the x-values where the curves intersect. Solve f(x) = g(x) for x.
Some regions require two or more integrals because the "top" function changes. For example, the area bounded by y = x, y = 4 – x, and y = 4 – x/3 might need to be split at the point where one boundary hands off to another.
When curves are given as x = f(y) and x = g(y), or when integrating in x would require splitting but integrating in y would not, use:
Area = ∫ (from c to d) |f(y) – g(y)| dy
where c and d are the y-coordinates of the intersection points. "Right minus left" replaces "top minus bottom."
Use whichever direction gives a single integral rather than multiple. If one curve is not a function of x (e.g. y² = 2x + 6 gives two branches), integrating with respect to y often avoids the split.
∫ f(g(x)) g'(x) dx = ∫ f(u) du, with u = g(x)
∫ (from a to b) f(g(x)) g'(x) dx = ∫ (from g(a) to g(b)) f(u) du
f even on [–a, a]: ∫ f dx = 2 ∫ (from 0 to a) f dx
f odd on [–a, a]: ∫ f dx = 0
x-direction: Area = ∫ (from a to b) |f(x) – g(x)| dx
y-direction: Area = ∫ (from c to d) |f(y) – g(y)| dy
Area between curves comes up when comparing two quantities over time: the area between a revenue curve and a cost curve gives total profit (or loss) over a period. In physics, the area between two velocity curves gives the difference in displacements.
Students forget to change the limits when doing u-substitution on a definite integral. If you change to u, the limits must change too; if you substitute back to x, the limits stay as they were.
Students pick u but forget to check that du (or a constant multiple of it) is present in the integrand. If du is missing entirely, the substitution will not work.
When finding area between curves, students sometimes subtract in the wrong order and get a negative result. Area is always non-negative. If your answer is negative, you subtracted the top from the bottom instead of bottom from top.
Students assume they can always integrate in x. Some problems are far easier in y. Always sketch the region first and decide.
Students apply the even/odd shortcut on intervals that are not symmetric about the origin. The formulas require limits of the form [–a, a].
⚠️ U-substitution is the most frequently tested integration technique in Calculus I. Expect two or three problems on any exam from this point on.
⚠️ Changing the limits of integration on a definite integral u-sub is explicitly practised on WeBWorK and tested on exams.
⚠️ Area between curves: you will be expected to find intersection points, set up the correct integral(s), and evaluate. Sketch the region.
⚠️ Even/odd trick: a quick way to simplify on symmetric intervals. Recognise odd integrands on [–a, a] instantly.
⚠️ Choosing x-direction vs. y-direction for area: exams sometimes give a region where x requires two integrals but y requires one (or vice versa). Know both approaches.
If u = x³ + 1, what is du? (3x² dx)
True or False: tan²(x) sec²(x) is an odd function. (False, it is even: both tan² and sec² are even.)
Fill in the blank: The area between y = 5 – x² and y = 1 is found by integrating from x = ____ to x = ____. (−2 to 2)
True or False: When doing u-sub on a definite integral, you must substitute back to x before evaluating. (False, you can change the limits instead.)
Fill in the blank: ∫ (from –a to a) xµ dx = ____. (0, because x⁵ is odd.)
Q: Evaluate ∫ x² √(x³ + 1) dx.
A: Let u = x³ + 1, du = 3x² dx, so x² dx = du/3. The integral becomes (1/3) ∫ √u du = (1/3)(2/3)u^(3/2) + C = (2/9)(x³ + 1)^(3/2) + C.
Q: Evaluate ∫ (from 0 to 1) ³√(1 + 7x) dx by changing limits.
A: Let u = 1 + 7x, du = 7 dx. When x = 0, u = 1. When x = 1, u = 8. Integral becomes (1/7) ∫ (from 1 to 8) u^(1/3) du = (1/7)(3/4)u^(4/3) |_1^8 = (3/28)(8^(4/3) – 1) = (3/28)(16 – 1) = 45/28.
Q: Find the area bounded between y = 1 and y = 5 – x².
A: Set 5 – x² = 1, so x² = 4, x = ±2. Top function is 5 – x², bottom is 1. Area = ∫ (from −2 to 2) [(5 – x²) – 1] dx = ∫ (from −2 to 2) (4 – x²) dx. The integrand is even, so = 2 ∫ (from 0 to 2) (4 – x²) dx = 2[4x – x³/3] from 0 to 2 = 2(8 – 8/3) = 2(16/3) = 32/3.
Q: Find the area enclosed by y = x – 1 and y² = 2x + 6.
A: Integrating in y is easier. Rewrite: x = y + 1 and x = (y² – 6)/2. Find intersections: y + 1 = (y² – 6)/2, so 2y + 2 = y² – 6, y² – 2y – 8 = 0, (y – 4)(y + 2) = 0, y = −2 and y = 4. Right curve: x = y + 1. Left curve: x = (y² – 6)/2. Area = ∫ (from −2 to 4) [(y + 1) – (y² – 6)/2] dy = 18.
Q: If f is continuous and ∫ (from 0 to 9) f(x) dx = 4, find ∫ (from 0 to 3) x f(x²) dx.
A: Let u = x², du = 2x dx. When x = 0, u = 0. When x = 3, u = 9. Integral becomes (1/2) ∫ (from 0 to 9) f(u) du = (1/2)(4) = 2. Answer: C.
U-substitution is the starting point for more advanced techniques covered in Calculus II: integration by parts, trigonometric substitution, and partial fractions all build on the same "transform and simplify" idea. Area between curves generalises to volumes of revolution (Ch. 6 in most textbooks) by rotating the enclosed region around an axis.
u-substitution, substitution rule, change of variable, change of limits, even function, odd function, symmetric integral, area between curves, area enclosed by curves, top minus bottom, right minus left, integrating with respect to y, intersection points, MTH 132, Calculus I, MSU, Michigan State, Chapter 4 Section 5, Chapter 5 Section 1