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Q: What does H2 Maths Notes (JC 1-2): 5.4) Definite Integrals cover?A: Area, volume, and mean-value interpretations for H2 Maths Topic 5.4.Before you revise Fundamental Theorem If
F ′ ( x ) = f ( x ) F'(x) = f(x) F ′ ( x ) = f ( x ) , then
∫ a b f ( x ) d x = F ( b ) − F ( a ) \int_a^b f(x) \space dx = F(b) - F(a) ∫ a b f ( x ) d x = F ( b ) − F ( a ) Part 21 of 32
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Properties: linearity, additivity over intervals, and reversal of limits introducing a negative sign.
Example -- Using an antiderivative
∫ 1 4 ( 3 x 2 − 2 x + 1 ) , d x = [ x 3 − x 2 + x ] 1 4 = ( 64 − 16 + 4 ) − ( 1 − 1 + 1 ) = 50.
\int_1^4 (3x^2 - 2x + 1) , dx = \left[ x^3 - x^2 + x \right]_1^4 = (64 - 16 + 4) - (1 - 1 + 1) = 50.
∫ 1 4 ( 3 x 2 − 2 x + 1 ) , d x = [ x 3 − x 2 + x ] 1 4 = ( 64 − 16 + 4 ) − ( 1 − 1 + 1 ) = 50.
Areas Between Curves For vertical slices:
∫ a b ( y top − y bottom ) d x \int_a^b \left( y_\text{top} - y_\text{bottom} \right) \space dx ∫ a b ( y top − y bottom ) d x .
For horizontal slices:
∫ c d ( x right − x left ) d y \int_c^d \left( x_\text{right} - x_\text{left} \right) \space dy ∫ c d ( x right − x left ) d y Determine intersection points by solving
y top = y bottom y_\text{top} = y_\text{bottom} y top = y bottom .
Find area between y = x 2 y = x^2 y = x 2 y = 2 x + 3 y = 2x + 3 y = 2 x + 3
Intersections solve
x 2 = 2 x + 3 x^2 = 2x + 3 x 2 = 2 x + 3 ⇒
x = − 1 , 3 x = -1, 3 x = − 1 , 3 .
Area
= ∫ − 1 3 ( 2 x + 3 − x 2 ) d x = [ x 2 + 3 x − x 3 3 ] − 1 3 = 32 3 = \int_{-1}^3 \left(2x + 3 - x^2\right) dx = \biggl[ x^2 + 3x - \dfrac{x^3}{3} \biggr]_{-1}^3 = \dfrac{32}{3} = ∫ − 1 3 ( 2 x + 3 − x 2 ) d x = [ x 2 + 3 x − 3 x 3 ] − 1 3 = 3 32 .
Volumes of Revolution About x-axis:
V = π ∫ a b y 2 d x V = \pi \int_a^b y^2 \space dx V = π ∫ a b y 2 d x .
About y-axis:
V = π ∫ c d x 2 d y V = \pi \int_c^d x^2 \space dy V = π ∫ c d x 2 d y .
Shell method (if convenient):
V = 2 π ∫ a b x y d x V = 2\pi \int_a^b x y \space dx V = 2 π ∫ a b x y d x .
Region under y = x y = \sqrt{x} y = x x = 0 x = 0 x = 0 x = 4 x = 4 x = 4
V = π ∫ 0 4 x d x = π [ x 2 2 ] 0 4 = 8 π .
V = \pi \int_0^4 x \space dx = \pi \biggl[ \dfrac{x^2}{2} \biggr]_0^4 = 8\pi.
V = π ∫ 0 4 x d x = π [ 2 x 2 ] 0 4 = 8 π .
Example -- Shell method about y-axis
Revolve the region bounded by y = 2 − x y = 2 - x y = 2 − x x = 0 x = 0 x = 0
V = 2 π ∫ 0 2 x ( 2 − x ) , d x = 2 π [ x 2 − x 3 3 ] 0 2 = 2 π ( 4 − 8 3 ) = 16 π 3 .
V = 2\pi \int_0^2 x (2 - x) , dx = 2\pi \left[ x^2 - \dfrac{x^3}{3} \right]_0^2 = 2\pi \left( 4 - \dfrac{8}{3} \right) = \dfrac{16\pi}{3}.
V = 2 π ∫ 0 2 x ( 2 − x ) , d x = 2 π [ x 2 − 3 x 3 ] 0 2 = 2 π ( 4 − 3 8 ) = 3 16 π .
Mean Value and Average Example -- Average height
The mean value of f ( x ) = sin x f(x) = \sin x f ( x ) = sin x [ 0 , π ] [0, \pi] [ 0 , π ]
f ˉ = 1 π − 0 ∫ 0 π sin x , d x = 1 π [ − cos x ] 0 π = 2 π .
\bar{f} = \dfrac{1}{\pi - 0} \int_0^\pi \sin x , dx = \dfrac{1}{\pi} \left[ -\cos x \right]_0^\pi = \dfrac{2}{\pi}.
f ˉ = π − 0 1 ∫ 0 π sin x , d x = π 1 [ − cos x ] 0 π = π 2 .
Integration with Modulus or Piecewise Functions Evaluate ∫ − 2 3 ∣ x − 1 ∣ d x \int_{-2}^3 \lvert x - 1 \rvert dx ∫ − 2 3 ∣ x − 1 ∣ d x
∫ − 2 1 ( 1 − x ) d x + ∫ 1 3 ( x − 1 ) d x = [ x − x 2 2 ] − 2 1 + [ x 2 2 − x ] 1 3 = 13 2 .
\int_{-2}^1 (1 - x) dx + \int_1^3 (x - 1) dx = \biggl[ x - \dfrac{x^2}{2} \biggr]_{-2}^1 + \biggl[ \dfrac{x^2}{2} - x \biggr]_1^3 = \dfrac{13}{2}.
∫ − 2 1 ( 1 − x ) d x + ∫ 1 3 ( x − 1 ) d x = [ x − 2 x 2 ] − 2 1 + [ 2 x 2 − x ] 1 3 = 2 13 .
Improper Definite Integrals
Infinite limits For
∫ a ∞ f ( x ) , d x \int_a^\infty f(x) , dx ∫ a ∞ f ( x ) , d x , evaluate
lim t → ∞ ∫ a t f ( x ) d x \lim_{t \to \infty} \int_a^t f(x) \space dx lim t → ∞ ∫ a t f ( x ) d x .
For
∫ − ∞ b f ( x ) , d x \int_{-\infty}^b f(x) , dx ∫ − ∞ b f ( x ) , d x , use
lim t → − ∞ ∫ t b f ( x ) d x \lim_{t \to -\infty} \int_t^b f(x) \space dx lim t → − ∞ ∫ t b f ( x ) d x ∫ 1 ∞ 1 x 2 d x = lim t → ∞ [ − 1 x ] 1 t = lim t → ∞ ( − 1 t + 1 ) = 1.
\int_1^\infty \dfrac{1}{x^2} \space dx = \lim_{t \to \infty} \left[ -\dfrac{1}{x} \right]_1^t = \lim_{t \to \infty} \left( -\dfrac{1}{t} + 1 \right) = 1.
∫ 1 ∞ x 2 1 d x = t → ∞ lim [ − x 1 ] 1 t = t → ∞ lim ( − t 1 + 1 ) = 1.
Vertical asymptotes or interior discontinuities Example -- Endpoint asymptote
∫ 0 1 1 x d x = lim ε → 0 + ∫ ε 1 x − 1 / 2 d x = lim ε → 0 + [ 2 x 1 / 2 ] ε 1 = 2.
\int_0^1 \dfrac{1}{\sqrt{x}} \space dx = \lim_{\varepsilon \to 0^+} \int_{\varepsilon}^1 x^{-1/2} \space dx = \lim_{\varepsilon \to 0^+} \left[ 2x^{1/2} \right]_{\varepsilon}^1 = 2.
∫ 0 1 x 1 d x = ε → 0 + lim ∫ ε 1 x − 1/2 d x = ε → 0 + lim [ 2 x 1/2 ] ε 1 = 2.
Example -- Interior discontinuity
∫ − 1 2 1 ( x − 1 ) 2 , d x = lim t → 1 − ∫ − 1 t 1 ( x − 1 ) 2 , d x + lim s → 1 + ∫ s 2 1 ( x − 1 ) 2 , d x
\int_{-1}^2 \dfrac{1}{(x - 1)^2} , dx = \lim_{t \to 1^-} \int_{-1}^t \dfrac{1}{(x - 1)^2} , dx + \lim_{s \to 1^+} \int_s^2 \dfrac{1}{(x - 1)^2} , dx
∫ − 1 2 ( x − 1 ) 2 1 , d x = t → 1 − lim ∫ − 1 t ( x − 1 ) 2 1 , d x + s → 1 + lim ∫ s 2 ( x − 1 ) 2 1 , d x
Each limit diverges, so the original integral diverges.
Convergence quick checks
Calculator Workflow Use GC definite integral function to verify numeric value after completing manual steps.
When dealing with piecewise functions, evaluate each segment separately on the GC as a check.
Document the command (e.g. ∫(2x+3-x^2,x,-1,3)) in working if used.
Exam Watch Points Always sketch the region and label axes/limits.
For revolutions, specify axis clearly and include
π \pi π factor.
State units (square or cubic) when context demands.
Justify convergence when evaluating improper integrals.
Practice Quiz Verify that you can manage area, volume, and improper integral workflows with precise limit handling.
Quick Revision Checklist Compute areas between curves accurately with correct limits and integrands.
Derive volumes of revolution using shell or disk method as appropriate.
Handle modulus and piecewise integrals by splitting at critical points.
Evaluate improper integrals with proper limit notation and convergence checks.
H2 Maths Notes (JC 1-2): 5.4) Definite Integrals
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. If both tails are infinite, split at a convenient point and handle each limit separately.
converges.
