Second fundamental theorem of calculus

The statement. Let $f$ be continuous on the closed interval $[a,b]$, and suppose $F$ is any function such that $F'(x) = f(x)$ for every $x \in [a,b]$ — that is, $F$ is an antiderivative (indefinite integral) of $f$. The second fundamental theorem of calculus says

where: $f$ is the integrand, continuous throughout $[a,b]$; $F$ is any one antiderivative of $f$, obtained from the standard list of indefinite integrals; $a$ and $b$ are the lower and upper limits; and $F(b) - F(a)$ is written compactly as $\big[F(x)\big]_a^b$.

Why any antiderivative works. Two antiderivatives of the same $f$ can only differ by a constant $C$ — that is a consequence of the mean value theorem you already accept. So if you pick $F(x) + C$ instead of $F(x)$, the theorem still gives the same number:

The constant cancels automatically. This is exactly why you never write "$+C$" once you switch from an indefinite integral to a definite one — it would vanish in the subtraction anyway, so there is no point carrying it.

The three-step recipe. In practice, evaluating $\int_a^b f(x)\,dx$ is always the same three moves. First, find one antiderivative $F(x)$ of $f(x)$ using the standard forms — the power rule, $\int e^x\,dx = e^x$, $\int \frac{1}{x}\,dx = \ln|x|$, the trigonometric integrals, and forms like $\int \frac{dx}{\sqrt{a^2-x^2}} = \sin^{-1}\frac{x}{a}$. Second, write the difference $F(b) - F(a)$, upper limit minus lower limit, in that order. Third, simplify the resulting number. Nothing about Riemann sums or rectangles is needed at this stage — the theorem has already done that work once, permanently, for every continuous $f$.

Continuity is the fine print. The theorem requires $f$ to be continuous on the whole closed interval $[a,b]$, not just at isolated points. If $f$ has a break — a vertical asymptote, an undefined point — anywhere inside $[a,b]$, the theorem cannot be applied across that break without extra care (splitting the interval, or recognising the integral may not exist at all). This single continuity condition is what separates a legitimate application of the theorem from an invalid one. The figure below shows the two pictures the theorem connects: the definite integral as the shaded area under the curve, computed geometrically, versus the same value obtained algebraically as $F(b) - F(a)$ from any antiderivative.

Evaluate $\displaystyle\int_0^{\pi/2}\cos x\,dx$, and then use the same method to evaluate $\displaystyle\int_1^2 \frac{dx}{x}$.

Step 1. For the first integral, recall $\dfrac{d}{dx}(\sin x) = \cos x$, so $F(x) = \sin x$ is an antiderivative of $\cos x$, and $\cos x$ is continuous on all of $[0,\pi/2]$.

Step 2. Apply the theorem: $\displaystyle\int_0^{\pi/2}\cos x\,dx = F\!\left(\frac{\pi}{2}\right) - F(0) = \sin\frac{\pi}{2} - \sin 0 = 1 - 0 = 1$.

Step 3. For the second integral, the antiderivative is $F(x) = \ln|x|$, valid and continuous throughout $[1,2]$ since $x$ never touches $0$ there. Applying the theorem, $\displaystyle\int_1^2 \frac{dx}{x} = F(2) - F(1) = \ln 2 - \ln 1 = \ln 2 - 0 = \ln 2$.

Step 4. Both answers follow the identical pattern — antiderivative, evaluate at the limits, subtract — regardless of whether the antiderivative is trigonometric or logarithmic. Final answers: $\int_0^{\pi/2}\cos x\,dx = \mathbf{1}$ and $\int_1^2 \frac{dx}{x} = \boldsymbol{\ln 2}$.

Competitive stretch

The same recipe extends immediately to the standard inverse-trigonometric forms. For $\displaystyle\int_0^1 \frac{dx}{1+x^2}$, the antiderivative is $\tan^{-1}x$, so the integral is $\big[\tan^{-1}x\big]_0^1 = \tan^{-1}1 - \tan^{-1}0 = \frac{\pi}{4} - 0 = \frac{\pi}{4}$. For $\displaystyle\int_0^{1/2} \frac{dx}{\sqrt{1-x^2}}$, the antiderivative is $\sin^{-1}x$, giving $\big[\sin^{-1}x\big]_0^{1/2} = \sin^{-1}\frac12 - \sin^{-1}0 = \frac{\pi}{6} - 0 = \frac{\pi}{6}$. Nothing new is invented here — only the entry from the standard list changes.

• Writing "$+C$" in a definite integral and trying to evaluate it — the constant always cancels in $F(b) - F(a)$, so simply omit it once limits are attached.
• Subtracting in the wrong order — the theorem is $F(b) - F(a)$, upper minus lower, never the reverse; swapping the order flips the sign of the final answer.
• Applying the theorem across a discontinuity — if $f$ is undefined or discontinuous anywhere inside $[a,b]$ (for example $f(x)=\frac{1}{x^2}$ on an interval containing $x=0$), the theorem cannot be used blindly; the continuity hypothesis must be checked first.
• Confusing this with the first fundamental theorem ($\frac{d}{dx}\int_a^x f(t)\,dt = f(x)$, a statement about differentiating an integral) — the second theorem is the one that evaluates a definite integral numerically, and it is this one tested throughout the Integrals chapter.

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