Skill: integration-techniques

Integration Techniques & Applications

Beyond the power rule: u-substitution runs the chain rule backwards, integration by parts trades a hard integral for an easier one, and both unlock area between curves and volumes of revolution.

1 u-substitution: the chain rule in reverse

Spot an inner function whose derivative is also hanging around in the integrand. Set u equal to it, rewrite everything in u, integrate, and substitute back.

u-substitution in 3 steps

1. u = inner 2. du = u′ dx (find it in the integrand) 3. ∫ f(u) du, then back-substitute

A good u makes du appear (up to a constant) in the integrand. For definite integrals, either change the bounds to u-bounds or back-substitute before evaluating.

Example ∫ 6x(x² + 1)³ dx

u = x² + 1, so du = 2x dx.
Rewrite: ∫ 6x(x²+1)³ dx = ∫ 3u³ du.
Integrate: 3u⁴/4 + C; back-substitute: (3/4)(x² + 1)⁴ + C.

Example ∫₀¹ 2x(x² + 1)² dx

u = x² + 1, du = 2x dx; when x = 0, u = 1; when x = 1, u = 2.
∫₁² u² du = [u³/3]₁² = 8/3 − 1/3 = 7/3.

2 Integration by parts: trade hard for easy

For products where substitution fails — typically x times something — the product rule run backwards trades ∫ u dv for the (hopefully easier) ∫ v du.

Integration by parts

∫ u dv = uv − ∫ v du

Choose u by LIATE priority: Log, Inverse trig, Algebraic (x, x²), Trig, Exponential. u should get simpler when differentiated.

Example ∫₀¹ x·e^x dx

Let u = x, dv = e^x dx; then du = dx, v = e^x.
∫₀¹ xe^x dx = [xe^x]₀¹ − ∫₀¹ e^x dx.
= (e − 0) − (e − 1) = 1.

3 Area between curves & volumes

Integrals measure between, not just under: top minus bottom, spun around an axis with disks.

Area between curves

A = ∫_a^b (top − bottom) dx

Find where the curves cross — those are your bounds a and b.

Volume by disks

V = ∫_a^b πR² dx

R is the radius: the distance from the region to the axis of rotation. Draw the region and the axis first.

Example Area between y = x and y = x²

Cross at x = 0 and x = 1; on [0, 1], x ≥ x² (top is y = x).
A = ∫₀¹ (x − x²) dx = [x²/2 − x³/3]₀¹ = 1/6.

Example Volume: y = x² on [0, 1] about the x-axis

Disks of radius R = x²: V = ∫₀¹ π(x²)² dx = π[x⁵/5]₀¹ = π/5.

Area between curves: integrate (top − bottom) between the crossing points.

4 Common mistakes

Three traps, each with the wrong version and the fix.

1. Choosing u so du doesn’t appear

Wrong: for ∫ x·e^x² dx, setting u = x. Right: u = x² gives du = 2x dx — the x is already in the integrand. Good u-substitutions make du (up to a constant) visible.

2. Forgetting to change bounds (or back-substitute)

Wrong: ∫₀¹ 2x(x²+1)² dx = ∫₀¹ u² du. Right: new variable, new bounds: u goes 1 → 2, so ∫₁² u² du. Or back-substitute before using 0 and 1.

3. Wrong radius in volumes

Wrong: revolving y = x² about the x-axis with R = x. Right: R is the distance from the region to the axis: here R = x² (the y-value), so V = π∫(x²)² dx. Draw it first.

Key vocabulary

u-substitution — u = inner function; rewrites a chained integrand as ∫ f(u) du.
Integration by parts — ∫ u dv = uv − ∫ v du; pick u by LIATE.
LIATE — Log, Inverse trig, Algebraic, Trig, Exponential: priority order for choosing u.
Area between curves — ∫_a^b (top − bottom) dx.
Disk method — V = ∫ πR² dx; washers when there’s a hole (πR² − πr²).

5 Quick checks

Cover the answers, decide, then reveal.

1. For ∫ x(x² + 5)⁴ dx, what are u and du?

u = x² + 5, du = 2x dx. The x in the integrand is (half of) du — that’s the signal.

2. For ∫ x·cos x dx, which choice follows LIATE?

u = x, dv = cos x dx. Algebraic beats trig in LIATE, and du = dx is simpler than x.

3. y = x³ on [0, 2] revolved about the x-axis: what is R?

R = x³. The radius is the y-value of the curve: V = π∫₀² x⁶ dx.

Practice now Try the substitution stepper Open the area/volume lab