Difficulty: Intermediate | Prerequisites: U-substitution, integration by parts, basic graphing.
This set of notes covers two applications that tie together everything from u-substitution and integration by parts: evaluating definite integrals that require these techniques, and using definite integrals to find the area enclosed between two curves. Both are standard exam material in Calculus II. If you are comfortable with the mechanics of substitution and by parts, this is where you put them to work on concrete, geometric problems.
Definite integrals with by parts follow the same formula but you evaluate the boundary term uv at both limits. To find the area between two curves, identify which function is on top, subtract the bottom from the top, and integrate over the interval where they intersect.
Definite integral
An integral with specified upper and lower limits that evaluates to a number rather than a family of functions. In simple terms, it gives you the net signed area under a curve between two points.
Area between curves
The region enclosed between two functions f(x) and g(x) over an interval [a, b]. You compute it as ∫ₐᵇ (top function - bottom function) dx. Think of it as stacking thin vertical slices and adding up their heights.
Intersection points
The x-values where f(x) = g(x). These define the limits of integration when finding the area between two curves. In simple terms, they are where the two curves cross, marking where the enclosed region begins and ends.
Top function / bottom function
Over a given interval, the top function is whichever has the larger y-value. The area formula always subtracts the lower from the higher. If the curves swap positions, you must split the integral at the crossing point.
The definite-integral version of the by-parts formula is:
\int_a^b u\, dv = [uv]_a^b - \int_a^b v\, duThe boundary term [uv]ₐᵇ means you evaluate uv at x = b and subtract its value at x = a, then deal with the remaining integral.
The lecture works through ∫₀^π sin(x)·sin(x) dx using by parts: u = sin x, dv = sin x dx, so du = cos x dx and v = -cos x.
This produces the original integral on the right side (a cyclic situation), and solving algebraically gives:
\int_0^\pi \sin^2(x)\, dx = \frac{\pi}{2}The same result follows from the identity sin²x = (1 - cos 2x)/2, but the by-parts route is instructive.
Set f(x) = g(x) and solve for the intersection points. These become your limits of integration.
On the interval between intersections, determine which function is on top (test a sample x-value, or sketch).
Integrate (top - bottom) dx from the left intersection to the right.
If the curves cross more than once, split into sub-intervals and handle each separately.
The lecture finds the area between y = x² ln(x) and y = 4 ln(x).
Setting them equal: x² ln(x) = 4 ln(x), so ln(x)(x² - 4) = 0. This gives x = 1 (from ln(x) = 0) and x = 2 (from x² - 4 = 0; only the positive root applies because ln(x) is undefined for x ≤ 0).
On [1, 2], testing shows 4 ln(x) ≥ x² ln(x), so the area is:
A = \int_1^2 \ln(x)(4 - x^2)\, dxThis itself requires integration by parts with u = ln(x) and dv = (4 - x²) dx, which the lecture evaluates to:
A = \frac{6 + 16\ln(2)}{3} - \frac{29}{9} = \frac{29 - 48\ln(2)}{9}(The exact simplification is detailed in the lecture; the important takeaway is the method, not memorising the arithmetic.)
Use by parts with u = x, dv = sin(x)cos(x) dx.
To integrate dv, use the substitution w = sin x, dw = cos x dx, giving ∫ sin x cos x dx = sin²x / 2. So v = sin²x / 2.
Apply the by-parts formula:
\left[\frac{x\sin^2 x}{2}\right]_0^\pi - \frac{1}{2}\int_0^\pi \sin^2(x)\, dxThe boundary term evaluates to 0 (since sin²(π) = sin²(0) = 0). The remaining integral is (1/2)(π/2) = π/4, so:
\int_0^\pi x\sin(x)\cos(x)\, dx = -\frac{\pi}{4}Intersection points: x = 1 and x = 2 (as derived in the core content section above).
The integral ∫₁² ln(x)(4 - x²) dx is evaluated by parts with u = ln(x), dv = (x² - 4) dx. The lecture works through the algebra step by step, arriving at (29 - 48 ln 2)/9.
Students sometimes forget to evaluate the boundary term [uv]ₐᵇ in definite by-parts integrals. This term is not zero unless you check.
When finding the area between curves, students often subtract in the wrong order (bottom minus top), producing a negative answer. Area is always non-negative; if you get a negative number, check which curve is above.
Students sometimes assume the intersection points are obvious and skip the algebra. Always solve f(x) = g(x) properly; missing an intersection point means integrating over the wrong interval.
For curves involving ln(x), students forget that the domain is restricted to x > 0. Negative x-solutions to intersection equations must be discarded.
⚠️ "Find the area between these two curves" is a near-guaranteed exam question. The full workflow, from finding intersections to setting up and evaluating the integral, is tested end to end.
⚠️ Definite integrals using by parts appear frequently, often combined with trig identities (sin²x, cos²x problems).
⚠️ The ∫₀^π sin²x dx = π/2 result appears so often in Calculus II that it is worth memorising outright.
⚠️ Problems may require both by parts and u-substitution in the same integral. Be ready to switch tools mid-problem.
True or False: The area between two curves can be negative.
Answer: False. Area is always non-negative. If your integral gives a negative value, you have the subtraction order reversed.
Fill in the blank: To find where two curves intersect, set ______ and solve for x.
Answer: f(x) = g(x)
True or False: ∫₀^π sin²(x) dx = π/2.
Answer: True.
Fill in the blank: In the definite by-parts formula, the term [uv]ₐᵇ means ______.
Answer: u(b)·v(b) - u(a)·v(a)
True or False: When finding the area between y = x² ln(x) and y = 4 ln(x), x = -2 is a valid intersection point.
Answer: False. ln(x) is undefined for x ≤ 0, so negative roots are discarded.
Q: Find the area between y = x and y = x² on [0, 1].
A: On [0, 1], x ≥ x². Area = ∫₀¹ (x - x²) dx = [x²/2 - x³/3]₀¹ = 1/2 - 1/3 = 1/6.
Q: Evaluate ∫₀^π x cos(x) dx using integration by parts.
A: u = x, dv = cos x dx, du = dx, v = sin x. [x sin x]₀^π - ∫₀^π sin x dx = 0 - [-cos x]₀^π = -(−cos π + cos 0) = -(1 + 1) = -2.
Q: Two curves cross at x = 1, x = 3, and x = 5. How many integrals do you need to compute the total enclosed area?
A: Two. One integral from x = 1 to x = 3, and another from x = 3 to x = 5. In each sub-interval you must check which curve is on top, because they may swap at each crossing.
Q: Why must you discard x = -2 as an intersection of y = x² ln(x) and y = 4 ln(x)?
A: Because ln(x) is defined only for x > 0. Both curves require a positive x-value to exist.
Area-between-curves problems reappear in Calculus II when you study volumes of revolution (disc, washer and shell methods): the cross-sectional area you integrate to find volume is itself an area-between-curves calculation.
The identity sin²x = (1 - cos 2x)/2, used here to evaluate ∫ sin²x dx, is a special case of the power-reduction formulas that come up repeatedly in trig integrals.
Definite integrals with logarithmic and exponential integrands also appear in probability and statistics (expected value calculations) and in physics (work, centre of mass).
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