Integral Calculus (9hrs) · Engineering Mathematics - I

Leibnitz's rule

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Leibnitz's rule

1. Full Leibniz Rule — Variable Limits

For F(α)=∫a(α)b(α)f(x,α)dxF(α)= ∫_{a(α)}^{b(α)} f(x,α) dx where f,fαf, f_α continuous and a(α),b(α)a(α), b(α) differentiable:

dF/dα=∫a(α)b(α)∂f/∂α(x,α)dx+f(b(α),α)⋅b′(α)−f(a(α),α)⋅a′(α)dF/dα = ∫_{a(α)}^{b(α)} ∂f/∂α (x,α) dx + f(b(α),α)·b'(α) - f(a(α),α)·a'(α)

i.e. partial differentiation inside + boundary contributions from moving limits. For constants a′,b′=0a',b'=0 it reduces to previous topic.

Special case F(x)=∫a(x)b(x)f(t)dtF(x)= ∫_{a(x)}^{b(x)} f(t) dt (integrand independent of xx):

F′(x)=f(b(x))b′(x)−f(a(x))a′(x)F'(x)= f(b(x))b'(x)-f(a(x))a'(x)

is the second fundamental theorem combined with chain rule.

2. Worked Examples

(a) F(α)= ∫_{0}^{α} sin(α x)/x dx? Need handle. Simpler:
    F(α)= ∫_{0}^{α²} e^{α x} dx = [e^{α x}/α]_{0}^{α²}= (e^{α³}-1)/α
    Via Leibniz: f=e^{α x}, f_α= x e^{α x}, b=α² ⇒ dF/dα= ∫_{0}^{α²} x e^{α x}dx + e^{α·α²}·2α - e^{0}·0
    Differentiate direct (1+α³)... matches ✓

(b) F(x)= ∫_{x}^{x²} t³ dt = x^8/4 - x^4/4
    Leibniz: f=t³, a=x, b=x² ⇒ F'= (x²)³·2x - x³·1 =2x^7 - x^3 matches differentiate ✓

(c) Evaluate I(α)= ∫_{0}^{π} log(1+α cos x)/cos x?? Use Leibniz to get ODE for I(α)

(d) F(x)= ∫_{0}^{x} (x - t) f(t) dt ⇒ F' = ∫_{0}^{x} f(t) dt, F''=f(x) (differentiate twice) ✓

(e) Classic: d/dx ∫_{0}^{x} sin(t²) dt = sin(x²) ; d/dx ∫_{0}^{x²} sin(t²) dt = sin(x⁴)·2x

3. Repeated Differentiation

Apply again to get higher derivatives: e.g., $d²F/dα² = ∫ f_{αα} dx + 2 f(b)??... plus terms from limits. Remember to differentiate boundary terms as well.

4. Exam Points

  • State full Leibniz d/dα∫a(α)b(α)f(x,α)dxd/dα ∫_{a(α)}^{b(α)} f(x,α)dx — 5 marks, very frequent.
  • Distinguish from fixed-limit case — 3 marks.
  • Evaluate d/dx∫xx2et2dtd/dx ∫_{x}^{x²} e^{t²} dt, d/dx∫0x(x−t)sintdtd/dx ∫_{0}^{x} (x-t)sin t dt type — 5-7 marks, very frequent.
  • Use Leibniz to prove d/dx∫0xf(t)(x−t)n−1/(n−1)!dt=∫0xf(t)(x−t)n−2/(n−2)!dtd/dx ∫_{0}^{x} f(t)(x-t)^{n-1}/(n-1)! dt = ∫_{0}^{x} f(t)(x-t)^{n-2}/(n-2)! dt — 5 marks, frequent.
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