Fundamental Theorem of Calculus

FTC I

If $g$ is continuous on $[a,b]$ that is differentiable on $(a,b)$ and if $g'$ is integrable on $[a,b]$ then

\int_a^b g' = g(b) - g(a)

Integration by Parts

Suppose $u,v$ are continuous functions on $[a,b]$ that are differentiable on $(a,b)$ . If $u'$ and $v'$ are Riemann integrable on $[a,b]$ :

\int_a^b u(x)v'(x)\,\text{d}x + \int_a^b u'(x)v(x)\,\text{d}x = u(b)v(b) - u(a)v(a)

FTC II

Suppose $f$ is an Riemann integrable function on $[a,b]$ . For $x\in(a,b)$ .

F(x)=\int_a^x f(t)\,\text{d}t

$F(x)$ is uniformly continuous on $[a,b]$
$f$ is continuous at $x_0 \in (a,b) \implies F$ is differentiable and $F'(x_0) = f(x_0)$

Variable Limits

Suppose $f$ is Riemann integrable on an open interval $I$ containing the values of differentiable functions $a,b$ . Then:

\frac {\text{d}} {\text{d}x} \int_{a(x)}^{b(x)} f(t)\,\text{d}t = f(b(x))b'(x) - f(a(x))a'(x)

Change of Variable

Suppose $u$ is a differentiable function on an open interval $J$ such that $u'$ is continuous. Let $I$ be an open interval such that $\forall x \in J,\;u(x) \in I$ .

If $f$ is continuous on $I$ , then $f \circ u$ is continuous on $J$ and:

\int_a^b{(f\circ u)(x)\,u'(x)\,\text{d}x} = \int_{u(a)}^{u(b)} f(u)\,\text{d}u

Written by

Sahithyan K.

November 2, 2026 1 min read

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