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Concept 08 / 10Integral Calculus

Differentiation Under the Integral Sign and Integral-Defined Functions

The Leibniz rule for d/dx of an integral with variable limits, analysing functions defined by integrals (monotonicity, extrema, limits), L'Hopital with integrals, and solving equations such as f(x) = x + integral of f.

By the fundamental theorem, ddx∫axf(t)dt=f(x)\frac{d}{dx}\int_{a}^{x}f(t)dt = f(x) for continuous ff. Combined with the chain rule, this gives the Leibniz rule for variable limits: ddx∫ψ(x)ϕ(x)f(t) dt=f(ϕ(x))ϕ′(x)−f(ψ(x))ψ′(x)\frac{d}{dx}\int_{\psi(x)}^{\phi(x)}f(t)\,dt = f(\phi(x))\phi'(x) - f(\psi(x))\psi'(x). Reason: if FF is an antiderivative of ff, the integral equals F(ϕ(x))−F(ψ(x))F(\phi(x)) - F(\psi(x)); differentiate each term with the chain rule. The upper limit contributes with a plus sign and the lower limit with a minus sign. If the integrand also depends on xx (not just the limits), first try to pull the xx out. For ∫0x(x−t)f(t)dt=x∫0xf(t)dt−∫0xtf(t)dt\int_{0}^{x}(x - t)f(t)dt = x\int_{0}^{x}f(t)dt - \int_{0}^{x}tf(t)dt, differentiating with the product rule gives ∫0xf(t)dt+xf(x)−xf(x)=∫0xf(t)dt\int_{0}^{x}f(t)dt + xf(x) - xf(x) = \int_{0}^{x}f(t)dt. (The general rule for an integrand g(x,t)g(x, t) adds a term ∫∂g∂xdt\int\frac{\partial g}{\partial x}dt, which is occasionally useful for evaluating integrals by introducing a parameter.) Functions defined by integrals. A function like F(x)=∫0xf(t)dtF(x) = \int_{0}^{x}f(t)dt can be analysed without evaluating the integral. F′(x)=f(x)F'(x) = f(x) tells you where FF increases and decreases, and its critical points are the zeros of ff. F′′(x)=f′(x)F''(x) = f'(x) gives concavity. F(0)=0F(0) = 0 gives a reference value. For example, F(x)=∫0x(t−1)(t−2)e−t2dtF(x) = \int_{0}^{x}(t - 1)(t - 2)e^{-t^{2}}dt has a local maximum at x=1x = 1 and a local minimum at x=2x = 2, found without integrating. Limits involving integrals. A limit like lim⁡x→0∫0xsin⁡t2 dtx3\lim_{x \to 0}\frac{\int_{0}^{x}\sin t^{2}\,dt}{x^{3}} is of the form 00\frac{0}{0}, and L'Hopital's rule with the Leibniz rule gives lim⁡sin⁡x23x2=13\lim\frac{\sin x^{2}}{3x^{2}} = \frac{1}{3}. Integrals of the form ∫0x\int_{0}^{x} with a small upper limit can also be approximated by x⋅f(0)x \cdot f(0) or by integrating a series term by term. Integral equations. If a function is defined by an equation containing a definite integral of itself with constant limits, such as f(x)=x+∫01tf(t)dtf(x) = x + \int_{0}^{1}tf(t)dt, the integral is just an unknown constant kk. Write f(x)=x+kf(x) = x + k, substitute back into the definition of kk, and solve. If the limits are variable, such as f(x)=1+∫0xf(t)dtf(x) = 1 + \int_{0}^{x}f(t)dt, differentiate to get a differential equation (f′=ff' = f), and use the value at the lower limit (f(0)=1f(0) = 1) as the initial condition: here f(x)=exf(x) = e^{x}. Key Takeaways: - ddx∫ψϕf(t)dt=f(ϕ)ϕ′−f(ψ)ψ′\frac{d}{dx}\int_{\psi}^{\phi}f(t)dt = f(\phi)\phi' - f(\psi)\psi'. - Pull out factors of xx from the integrand before differentiating. - For F(x)=∫axfF(x) = \int_{a}^{x}f: F′=fF' = f, F(a)=0F(a) = 0; analyse FF through ff. - 00\frac{0}{0} limits with integrals: L'Hopital plus Leibniz. - Constant-limit integral equations: the integral is a constant kk; solve for kk. - Variable-limit integral equations: differentiate, then use the value at the lower limit. - Pitfall: forgetting the chain-rule factor ϕ′(x)\phi'(x) for a non-trivial upper limit. Brain-Teaser 1: Find ddx∫x2x3dtln⁡t\frac{d}{dx}\int_{x^{2}}^{x^{3}}\frac{dt}{\ln t} for x>1x > 1. Solution: By the Leibniz rule: 1ln⁡x3⋅3x2−1ln⁡x2⋅2x=3x23ln⁡x−2x2ln⁡x=x2−xln⁡x\frac{1}{\ln x^{3}} \cdot 3x^{2} - \frac{1}{\ln x^{2}} \cdot 2x = \frac{3x^{2}}{3\ln x} - \frac{2x}{2\ln x} = \frac{x^{2} - x}{\ln x}. The integrand 1ln⁡t\frac{1}{\ln t} has no elementary antiderivative, yet the derivative is elementary, which is exactly the power of the rule. Brain-Teaser 2: Find ff if f(x)=x2+∫01xt f(t) dtf(x) = x^{2} + \int_{0}^{1}xt\,f(t)\,dt. Solution: The integral is x∫01tf(t)dt=kxx\int_{0}^{1}tf(t)dt = kx with k=∫01tf(t)dtk = \int_{0}^{1}tf(t)dt a constant. So f(x)=x2+kxf(x) = x^{2} + kx. Substituting: k=∫01t(t2+kt)dt=14+k3k = \int_{0}^{1}t(t^{2} + kt)dt = \frac{1}{4} + \frac{k}{3}, so 2k3=14\frac{2k}{3} = \frac{1}{4} and k=38k = \frac{3}{8}. Hence f(x)=x2+38xf(x) = x^{2} + \frac{3}{8}x. Recognising the integral as a constant (times xx) turns the functional equation into a linear equation for one number. Brain-Teaser 3: Evaluate lim⁡x→0∫0x2sec⁡2t dtxsin⁡x\lim_{x \to 0}\frac{\int_{0}^{x^{2}}\sec^{2}t\,dt}{x\sin x}. Solution: Both numerator and denominator tend to 00. By L'Hopital and Leibniz: the numerator's derivative is sec⁡2(x2)⋅2x\sec^{2}(x^{2}) \cdot 2x and the denominator's is sin⁡x+xcos⁡x\sin x + x\cos x. So the limit is lim⁡2xsec⁡2x2sin⁡x+xcos⁡x=lim⁡2sec⁡2x2sin⁡xx+cos⁡x=21+1=1\lim\frac{2x\sec^{2}x^{2}}{\sin x + x\cos x} = \lim\frac{2\sec^{2}x^{2}}{\frac{\sin x}{x} + \cos x} = \frac{2}{1 + 1} = 1. Alternatively, for small xx the numerator is about x2x^{2} (since sec⁡2t≈1\sec^{2}t \approx 1) and the denominator about x2x^{2}, giving 11 at once.

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Differentiation Under the Integral Sign and Integral-Defined Functions — DPP

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