Calculus: Integration and Applications
What this domain tests
Antiderivatives from the standard table, integration by linear substitution and by parts, the fundamental theorem, definite integrals and their properties, and areas of plane regions - including regions between two curves and, where your syllabus includes it, volumes of revolution.
Core rules
- ∫x^n dx = x^(n+1)/(n+1) + c for n ≠ −1; ∫(1/x)dx = ln|x| + c; ∫e^x dx = e^x + c; ∫sin x dx = −cos x + c; ∫cos x dx = sin x + c; ∫dx/cos²x = tan x + c.
- Linear substitution: ∫f(ax + b)dx = (1/a)F(ax + b) + c. The factor 1/a is compulsory.
- By parts: ∫u·v′ dx = uv − ∫u′·v dx. Choose u as the factor that simplifies when differentiated - logarithm first, then power, then exponential or trigonometric.
- Fundamental theorem: ∫ from a to b of f = F(b) − F(a). If you substitute in a definite integral, change the limits to the new variable.
- Area between curves: ∫ from a to b of |f(x) − g(x)| dx, splitting the interval where f = g. The definite integral gives signed area; geometric area needs absolute values.
- Solid of revolution about the x-axis: V = π∫ from a to b of y² dx.
Worked example
Find ∫x ln x dx for x > 0. Take u = ln x and v′ = x, so u′ = 1/x and v = x²/2. Then
∫x ln x dx = (x²/2)ln x − ∫(x²/2)(1/x)dx = (x²/2)ln x − (1/2)∫x dx = (x²/2)ln x − x²/4 + c.
Check by differentiating: x ln x + x/2 − x/2 = x ln x, which is the integrand, so the antiderivative is correct. Note the ln x factor must be the part you differentiate; choosing u = x instead leads back to the original integral.
Traps that cost marks
- Omitting + c in an indefinite integral.
- Forgetting the 1/a in ∫f(ax + b)dx, e.g. writing ∫cos 2x dx = sin 2x.
- Leaving old limits after a substitution, or forgetting to convert back.
- Writing ln x instead of ln|x| when the integrand is 1/x on an interval containing negatives.
- "Integrating" a product term by term: ∫x·e^x dx ≠ (x²/2)·e^x.
- Reporting a negative number as an area - check where each curve lies.