Definite Integrals and the Fundamental Theorem

Two directions of one connection

The Fundamental Theorem of Calculus connects a limiting sum with a derivative. If $f$ is continuous on an interval and $A(x)=\int_a^x f(t)\,dt$ , then $A'(x)=f(x)$ . The new accumulation added by a short extension of the interval is approximately the local height times that extension.

Conversely, if $F'=f$ and $f$ is continuous on $[a,b]$ , then

\int_a^b f(x)\,dx=F(b)-F(a).

This evaluates the integral without constructing a sequence of partitions. The integration constant cancels in the endpoint difference.

Visual guide

VISUAL GUIDESigned accumulation can cancel

f(x) = x

For f(x) = x on [−1, 1], the negative triangular contribution is −1/2 and the positive contribution is +1/2. The definite integral is zero, although total geometric area is 1.

Worked example: evaluation and sign

For $\int_0^2(3x^2-1)\,dx$ , use $F=x^3-x$ . The value is $F(2)-F(0)=6$ . The integrand is initially negative and later positive, so the answer is net signed area rather than the sum of all geometric pieces.

Reversing bounds changes the sign: $\int_2^0(3x^2-1)dx=-6$ . Splitting at an intermediate point preserves the total: $\int_a^b f=\int_a^c f+\int_c^b f$ . These identities are helpful for splitting piecewise functions and checking signs.

Worked example: differentiate accumulation

Let $H(x)=\int_1^{x^2}\cos t\,dt$ . First regard $G(u)=\int_1^u\cos t\,dt$ , so $G'(u)=\cos u$ . Since $H=G(x^2)$ , the chain rule gives $H'(x)=2x\cos(x^2)$ .

If both bounds depend on $x$ ,

\frac{d}{dx}\int_{u(x)}^{v(x)}f(t)dt=f(v(x))v'(x)-f(u(x))u'(x).

This follows by splitting into two accumulations with the same fixed base point. If the integrand also explicitly depends on $x$ , a more general rule is needed; do not apply this formula without noticing that extra dependence.

Net change and averages

For a differentiable position with continuous velocity, $\int_a^b v(t)dt=s(b)-s(a)$ . The units are velocity times time, hence distance units. The average value of a continuous function is $(b-a)^{-1}\int_a^b f$ , which has the same units as the function itself.

Antiderivative evaluation across a singularity is invalid unless the integral is treated with the necessary one-sided limits. For example, blindly using $\ln|x|$ from $-1$ to $1$ does not establish an integral of $1/x$ through zero.

Practice

Evaluate $\int_1^3 2x\,dx$ .
Differentiate $\int_0^{\sin x}e^t\,dt$ .
Find the average value of $x^2$ on $[0,3]$ .

Show worked solutions

$[x^2]_1^3=9-1=8$ .
Evaluate the integrand at the upper bound and multiply by its derivative: $e^{\sin x}\cos x$ .
The integral is $9$ , and the interval length is $3$ , giving average value $3$ .