Fundamental theorem of calculus

The statement. For $f$ continuous on $[a,b]$, if $F'(x) = f(x)$ for every $x$ in $[a,b]$, then:

where: $f$ is the integrand, $F$ is any antiderivative of $f$ (i.e. $\int f(x)\,dx = F(x)+C$), and $a,b$ are the lower and upper limits of integration. The result is often abbreviated $\big[F(x)\big]_a^b = F(b)-F(a)$.

Why the constant $C$ never matters here. An indefinite integral is a whole family of functions, $F(x)+C$, since differentiating any constant gives zero. But in a definite integral the $C$ cancels automatically: $\big[F(x)+C\big]_a^b = (F(b)+C)-(F(a)+C) = F(b)-F(a)$. So you can drop $C$ the moment you switch from indefinite to definite integration — a definite integral is always a single number, never a family.

The differentiation form. The theorem also has a second face: differentiating the "area accumulated so far" gives back the original function,

where: the left side is the derivative, with respect to the upper limit $x$, of the area function built from $f$. ISC problems mainly use the evaluation form $F(b)-F(a)$, but this differentiation form is what makes the theorem "fundamental" — it shows integration and differentiation undo each other.

Why this saves so much work. Computing $\int_a^b f(x)\,dx$ directly from its limit-of-sums (Riemann-sum) definition means summing infinitely many thin rectangular strips and taking a limit as their width shrinks to zero — a genuinely hard calculation for anything beyond the simplest functions. The theorem replaces that entire limiting process with two substitutions: find any $F$, then compute $F(b)-F(a)$.

Continuity matters. The theorem as used at this level requires $f$ to be continuous on the closed interval $[a,b]$ — no jumps, no undefined points, no vertical asymptotes inside the interval. If the integrand has a discontinuity somewhere inside $[a,b]$, you cannot apply $F(b)-F(a)$ across the whole interval in one step; instead you split the integral at the discontinuity into two (or more) pieces, each continuous, and apply the theorem separately on each piece before adding the results. This restriction rarely bites in ISC problems, since most integrands there (polynomials, trigonometric, exponential, and their standard combinations) are continuous wherever they are defined, but it is worth checking, especially for integrands with a denominator that could vanish inside $[a,b]$.

Evaluate $\displaystyle\int_0^{\pi/2}\cos x\,dx$, then use the theorem on two more integrands that require prior techniques.

Step 1. Antiderivative: since $\dfrac{d}{dx}(\sin x) = \cos x$, take $F(x)=\sin x$. Then $\displaystyle\int_0^{\pi/2}\cos x\,dx = \big[\sin x\big]_0^{\pi/2} = \sin\dfrac{\pi}{2}-\sin 0 = 1-0=1$.

Step 2. Now try $\displaystyle\int_0^1 x\,e^{x^2}\,dx$, which needs substitution first. Let $t=x^2$, so $dt=2x\,dx$, i.e. $x\,dx=\dfrac{dt}{2}$; the limits change too, since $x=0\Rightarrow t=0$ and $x=1\Rightarrow t=1$. Then $\displaystyle\int_0^1 xe^{x^2}dx = \int_0^1 e^t\cdot\frac{dt}{2} = \frac12\big[e^t\big]_0^1 = \frac12(e-1)$. The theorem applies smoothly here, provided the limits are converted to the new variable $t$ as well.

Step 3. Finally $\displaystyle\int_0^1 x\,e^x\,dx$, which needs integration by parts: $\int xe^x\,dx = e^x(x-1)+C$. Applying the theorem, $\displaystyle\int_0^1 xe^x\,dx = \big[e^x(x-1)\big]_0^1 = e^1(0)-e^0(-1) = 0-(-1)=1$.

Final answers: $\int_0^{\pi/2}\cos x\,dx = 1$, $\int_0^1 xe^{x^2}dx = \dfrac{e-1}{2}$, and $\int_0^1 xe^x\,dx = 1$. All three use the same last move, $F(b)-F(a)$, regardless of which technique produced $F$.

Competitive stretch

For $\displaystyle\int_0^1 xe^{x^2}dx$, a common shortcut skips the explicit substitution and reads off the antiderivative directly: since $\dfrac{d}{dx}\left(\tfrac12 e^{x^2}\right) = x\,e^{x^2}$ by the chain rule, $F(x)=\tfrac12 e^{x^2}$ immediately, giving $F(1)-F(0) = \tfrac12(e-1)$ in one line — the same answer, reached by spotting the antiderivative pattern instead of substituting.

• Re-adding the constant $C$ in a definite integral — it is unnecessary, since it always cancels in $F(b)-F(a)$; carrying it around risks arithmetic slips.
• Forgetting to change the limits of integration after a substitution — either convert limits to the new variable throughout, or substitute back to the original variable first, but never mix the two approaches.
• Sign or arithmetic slips in $F(b)-F(a)$, especially when $F(a)$ is negative — always write both evaluations separately before subtracting.
• Assuming the theorem applies when $f$ is discontinuous somewhere in $[a,b]$, or confusing the definite-integral value with the antiderivative $F(x)$ itself — a jump or undefined point inside the interval means the integral must be split first, and a definite integral is a single number, not the function $F(x)$; writing "$\int_a^b f(x)\,dx = F(x)$" without evaluating at the limits is incomplete.

Ask the tutor: Want a walkthrough of how to correctly switch limits when substituting inside a definite integral, or which antiderivative technique to pick before applying the theorem?