OCR section 1.08 takes integration far past reversing the power rule. You need substitution, by parts, partial fractions, the trapezium rule, and differential equations solved by separating the variables.
Choosing the right technique is most of the battle — we will drill that. Work through each screen, answer the questions as you go and collect ⭐ stars. Press Start when you're ready.
The n = −1 case is exactly why ln exists — the power rule would divide by zero.
∫ from 1 to 3 of (2x + 1) dx = [x² + x] from 1 to 3
= (3² + 3) − (1² + 1) = 12 − 2 = 10.
Check geometrically: this is a trapezium under the line y = 2x + 1 from x = 1 to 3, with parallel sides 3 and 7 and width 2. Area = ½(3 + 7)(2) = 10. ✓
Area below the axis counts negative. If a curve crosses the x-axis inside your limits, a single integral gives the net signed area. To get the true total area you must split the integral at the root and add the magnitudes.
Substitution is the chain rule in reverse. Look for a function and its derivative both sitting in the integrand.
Let u = x² + 5. Then du/dx = 2x, so du = 2x dx — and 2x dx is exactly what is sitting there.
The integral becomes ∫ u⁶ du = u⁷/7 + c = (x² + 5)⁷/7 + c.
Check by differentiating: d/dx[(x²+5)⁷/7] = 7(x²+5)⁶(2x)/7 = 2x(x²+5)⁶. ✓
For a definite integral, change the limits too. If u = x² + 5 and x runs 0 → 2, then u runs 5 → 9. Change the limits and you never have to substitute back.
By parts is the product rule in reverse. Use it when the integrand is a product of two unrelated functions (so substitution will not work).
Take u = x (differentiates to 1 — simpler) and dv/dx = ex (so v = ex).
∫ x ex dx = x ex − ∫ ex dx = x ex − ex (+ c).
Evaluate from 0 to 1: at x = 1, e − e = 0. At x = 0, 0 − 1 = −1.
∴ the integral = 0 − (−1) = 1.
The special case: for ∫ ln x dx take u = ln x and dv/dx = 1. Then ∫ ln x dx = x ln x − ∫ x(1/x) dx = x ln x − x + c. Choosing u the other way round makes it worse, not better.
An integrand you cannot touch becomes two you can — split it first.
From the partial fractions work: (5x + 1)/((x − 1)(x + 2)) = 2/(x − 1) + 3/(x + 2).
∫ 2/(x − 1) dx + ∫ 3/(x + 2) dx = 2 ln|x − 1| + 3 ln|x + 2| + c.
Which can be tidied to ln[ (x − 1)² (x + 2)³ ] + c.
Keep the modulus signs. ∫1/(x − a) dx = ln|x − a| + c. Dropping the modulus is a mark lost whenever x < a.
A first-order separable differential equation can be split so that all the y terms sit with dy and all the x terms with dx — then integrate both sides.
Separate: (1/y) dy = 2x dx.
Integrate: ln|y| = x² + c.
∴ y = ex² + c = A ex², where A = ec.
Apply the condition y = 3 at x = 0: 3 = A e⁰ = A. So y = 3ex².
At x = 1: y = 3e = 8.15 (3 s.f.).
The general solution has an arbitrary constant; a particular solution uses a boundary condition to pin it down. Find the constant as soon as you have integrated — it is far easier before you rearrange.
When a function cannot be integrated exactly, approximate the area with trapezia.
h = (2 − 0)/4 = 0.5. Ordinates at x = 0, 0.5, 1, 1.5, 2:
y₀ = 1, y₁ = √1.25 = 1.11803, y₂ = √2 = 1.41421, y₃ = √3.25 = 1.80278, y₄ = √5 = 2.23607.
Sum of the middle ordinates = 1.11803 + 1.41421 + 1.80278 = 4.33502.
Estimate = (0.5/2)[1 + 2.23607 + 2(4.33502)] = 0.25 × 11.90612 = 2.977 (3 d.p.).
Over or under? The trapezium chords lie above a curve that bends upwards (convex), so the rule overestimates. √(x²+1) is convex here, and the exact value is 2.958 — our 2.977 is indeed too big. ✓
Tap an integral, then tap the technique you would use.
Tap an integral on the left, then its result on the right.
Standard: ∫xn dx = xn+1/(n+1) + c, except n = −1, which gives ln|x| + c
Area: area below the axis is negative — split the integral at the roots for total area
Substitution: the chain rule backwards — spot a function and its derivative; change the limits
By parts: ∫u(dv/dx)dx = uv − ∫v(du/dx)dx; pick u to be the part that simplifies
Partial fractions: split, then integrate to ln terms — keep the modulus signs
Trapezium: (h/2)[y₀ + yn + 2(middles)]; overestimates a convex curve
Differential equations: separate the variables, integrate, then use the boundary condition
That is OCR 1.08 — the largest toolkit in the Pure paper. Press Finish to see your score.
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