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Concept 06 / 10Integral Calculus

Definite Integrals: Limit of a Sum and the Fundamental Theorem of Calculus

The definite integral as a limit of Riemann sums and signed area, the fundamental theorem linking it to antiderivatives, evaluating limits of sums by converting them to integrals, and changing limits under substitution.

Figure 1
For a function ff on [a,b][a, b], divide the interval into nn equal parts of width Δx=b−an\Delta x = \frac{b - a}{n} and form the sum of rectangle areas ∑i=1nf(xi)Δx\sum_{i=1}^{n}f(x_{i})\Delta x, where xix_{i} is a point in the iith subinterval (often the right endpoint a+iΔxa + i\Delta x). As n→∞n \to \infty these sums approach a limit for every continuous ff, and that limit is the definite integral: ∫abf(x) dx=lim⁡n→∞∑i=1nf(a+ib−an)b−an\int_{a}^{b}f(x)\,dx = \lim_{n \to \infty}\sum_{i=1}^{n}f\left(a + i\frac{b - a}{n}\right)\frac{b - a}{n}. The figure shows the rectangles for n=8n = 8. Geometrically, the integral is the signed area between the curve and the xx-axis: regions above the axis count positive and regions below count negative. By convention ∫baf=−∫abf\int_{b}^{a}f = -\int_{a}^{b}f and ∫aaf=0\int_{a}^{a}f = 0. The fundamental theorem of calculus (FTC). Part 1: if ff is continuous and F(x)=∫axf(t) dtF(x) = \int_{a}^{x}f(t)\,dt, then F′(x)=f(x)F'(x) = f(x). Accumulated area grows at a rate equal to the current height. Part 2: if FF is any antiderivative of ff on [a,b][a, b], then ∫abf(x) dx=F(b)−F(a)\int_{a}^{b}f(x)\,dx = F(b) - F(a). Part 2 is what makes definite integrals computable: find an antiderivative and subtract. The constant CC cancels, which is why it is never written in definite integrals. The hypothesis of continuity on the whole interval is essential. ∫−11dxx2\int_{-1}^{1}\frac{dx}{x^{2}} is not [−1x]−11=−2\left[-\frac{1}{x}\right]_{-1}^{1} = -2; the integrand blows up at 00, and the (improper) integral actually diverges. A positive function cannot have a negative integral, which is a useful sanity check. Similarly, if the antiderivative has a jump in the interval (for example from a formula involving tan⁡−1\tan^{-1} of something that passes through infinity), split the interval at that point. Limit of a sum as an integral. JEE frequently reverses the definition. To evaluate lim⁡n→∞∑r=1n1nf(rn)\lim_{n \to \infty}\sum_{r=1}^{n}\frac{1}{n}f\left(\frac{r}{n}\right), recognise it as ∫01f(x) dx\int_{0}^{1}f(x)\,dx. The recipe: replace rn\frac{r}{n} by xx and 1n\frac{1}{n} by dxdx; the limits of xx are the limits of rn\frac{r}{n} as n→∞n \to \infty (the lower limit of rr divided by nn, and the upper limit of rr divided by nn). For instance, ∑r=12n\sum_{r=1}^{2n} gives xx from 00 to 22. The summand must first be rewritten as 1n\frac{1}{n} times a function of rn\frac{r}{n} alone. Substitution in definite integrals. When substituting u=g(x)u = g(x), change the limits to g(a)g(a) and g(b)g(b), and there is then no need to convert back to xx. The substitution must be valid (continuous and one-to-one) on the interval. Key Takeaways: - ∫abf=lim⁡∑f(xi)Δx\int_{a}^{b}f = \lim\sum f(x_{i})\Delta x; signed area. - FTC: ddx∫axf(t)dt=f(x)\frac{d}{dx}\int_{a}^{x}f(t)dt = f(x); ∫abf=F(b)−F(a)\int_{a}^{b}f = F(b) - F(a). - Check continuity on [a,b][a, b] before applying F(b)−F(a)F(b) - F(a). - lim⁡∑r=1n1nf(rn)=∫01f(x)dx\lim\sum_{r=1}^{n}\frac{1}{n}f(\frac{r}{n}) = \int_{0}^{1}f(x)dx; limits from rn\frac{r}{n} at the ends. - Change limits when substituting; do not revert to xx. - ∫ba=−∫ab\int_{b}^{a} = -\int_{a}^{b}. - Pitfall: applying FTC across a point where the integrand is infinite. Brain-Teaser 1: Evaluate lim⁡n→∞(1n+1+1n+2+⋯+12n)\lim_{n \to \infty}\left(\frac{1}{n + 1} + \frac{1}{n + 2} + \cdots + \frac{1}{2n}\right). Solution: The sum is ∑r=1n1n+r=∑r=1n1n⋅11+r/n\sum_{r=1}^{n}\frac{1}{n + r} = \sum_{r=1}^{n}\frac{1}{n} \cdot \frac{1}{1 + r/n}. With x=rnx = \frac{r}{n} running from 00 to 11, the limit is ∫01dx1+x=ln⁡2\int_{0}^{1}\frac{dx}{1 + x} = \ln 2. Factoring out 1n\frac{1}{n} so that everything else depends only on rn\frac{r}{n} is the essential step. Brain-Teaser 2: Evaluate ∫02x2 dx\int_{0}^{2}x^{2}\,dx directly from the limit of a sum. Solution: With Δx=2n\Delta x = \frac{2}{n} and right endpoints xi=2inx_{i} = \frac{2i}{n}: ∑i=1n(2in)22n=8n3∑i2=8n3⋅n(n+1)(2n+1)6\sum_{i=1}^{n}\left(\frac{2i}{n}\right)^{2}\frac{2}{n} = \frac{8}{n^{3}}\sum i^{2} = \frac{8}{n^{3}} \cdot \frac{n(n + 1)(2n + 1)}{6}. As n→∞n \to \infty this tends to 8×26=83\frac{8 \times 2}{6} = \frac{8}{3}. The FTC gives [x33]02=83[\frac{x^{3}}{3}]_{0}^{2} = \frac{8}{3} immediately, which shows how much work the theorem saves. Brain-Teaser 3: A student computes ∫0πdx1+2sin⁡2x\int_{0}^{\pi}\frac{dx}{1 + 2\sin^{2}x} by dividing by cos⁡2x\cos^{2}x and substituting u=tan⁡xu = \tan x, getting ∫00…=0\int_{0}^{0}\ldots = 0. What went wrong, and what is the correct value? Solution: The substitution u=tan⁡xu = \tan x is discontinuous at x=π2x = \frac{\pi}{2}, which lies inside [0,π][0, \pi], so the limits cannot simply be changed to tan⁡0=tan⁡π=0\tan 0 = \tan\pi = 0. The integrand is positive, so the answer cannot be 00. Split at π2\frac{\pi}{2}, or use symmetry: the integral equals 2∫0π/2dx1+2sin⁡2x2\int_{0}^{\pi/2}\frac{dx}{1 + 2\sin^{2}x}. On [0,π2)[0, \frac{\pi}{2}), dividing by cos⁡2x\cos^{2}x gives ∫0∞du1+3u2=13⋅π2\int_{0}^{\infty}\frac{du}{1 + 3u^{2}} = \frac{1}{\sqrt{3}} \cdot \frac{\pi}{2}. So the answer is π3\frac{\pi}{\sqrt{3}}. Every substitution must be continuous and monotonic on the whole interval.

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Definite Integrals: Limit of a Sum and the Fundamental Theorem of Calculus — DPP

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