Fundamental Theorem of Calculus

Fundamental Theorem of Calculus – HMC Calculus Tutorial

We are all used to evaluating definite integrals without giving the reason for the procedure much thought. The definite integral is defined not by our regular procedure but rather as a limit of Riemann sums. We often view the definite integral of a function as the area under the graph of the function between two limits. It is not intuitively clear, then, why we proceed as we do in computing definite integrals. The Fundamental Theorem of Calculus justifies our procedure of evaluating an antiderivative at the upper and lower limits of integration and taking the difference.

Let $f$ be continuous on $[a,b]$. If $F$ is any antiderivative for $f$ on $[a,b]$, then \[\int^b_a f(t)\, dt=F(b)-F(a).\]

Here’s a sketch of the proof, based on Salas and Hille’s Calculus: One Variable.

Let $\displaystyle G(x)=\int^x_a f(t)\, dt$.

Then it may be proven that $G(x)$ is an antiderivative for
$f$ on $[a,b]$.

Let $f$ be continuous on $[a,b]$. Then $F$ is an antiderivative for $f$ on $[a,b]$ if and only if $F$ is a continuous function on $[a,b]$ and $F'(x)=f(x)$ for all $x\in (a,b)$.

Let $F(x)$ be another antiderivative for $f$ on
$[a,b]$. Then $G(x)$ and $F(x)$ are continuous on $[a,b]$ and satisfy
$G'(x)=F'(x)=f(x)$ for all $x$ in $[a,b]$. It may be shown that
$F(x)$ and $G(x)$ differ only by a constant:
$G(x)=F(x)+C \quad\textrm{for some $C$ and all $x\in [a,b]$}$.
Now $G(a)=\int^a_a f(t)\, dt=0$, so $0=G(a)=F(a)+C$.
Then $C=-F(a)$, so $G(x)=F(x)-F(a).$
Letting $x=b$, $G(b)=F(b)-F(a)$,
so
$\int^b_a f(t)\, dt=F(b)-F(a)$.

Notation

We often write $\displaystyle \int^b_a \! f(t)\, dt=F(t)|^b_a$ or $\displaystyle \int^b_a \! f(t)\, dt=F(t)|^{t=b}_{t=a}$ to emphasize the variable with respect to which we are integrating.

Example

\begin{eqnarray*} \int^3_1 x\, dx&=& \left.\frac{x^2}{2}\right|^3_1\\ &=& \frac{3^2}{2}-\frac{1^2}{2}\\ &=& 4. \end{eqnarray*}

$\qquad$

Follow the image link for a complete description of the image

$Area_{shaded} = Area_{large \triangle} – Area_{small \triangle}$ $= \frac{1}{2}(3^2)-\frac{1}{2}(1^2) = 4$

If we had chosen a different antiderivative $\displaystyle \frac{x^2}{2}+C$, the outcome would have been identical: \begin{eqnarray*} \int^3_1 x\, dx=\left.\left(\frac{x^2}{2}+C\right)\right|^3_1&=&\left(\frac{9}{2}+C\right)- \left(\frac{1}{2}+C\right)\\ &=& \frac{9}{2}+C-\frac{1}{2}-C\\ &=&4. \end{eqnarray*}

Properties

$\displaystyle \int^a_a f(x)\, dx=0$.

\[\int^a_b f(x)\, dx=-\int^b_a f(x)\, dx\] Proof:

Let $F$ be an antiderivative for $f$ on $[a,b]$. Then \[\int^a_b f(x)\, dx=F(a)-F(b)=-(F(b)-F(a))=-\int^b_a f(x)\, dx,\] where the first and third equalities are applications of the Fundamental Theorem of Calculus.

\[\int^b_a [\alpha f(x)+\beta g(x)]\, dx=\alpha \int^b_a f(x)\, dx+\beta \int^b_a g(x)\, dx.\] Proof:

Let $F$ be an antiderivative for $f$ and $G$ be an antiderivative for $g$ on $[a,b]$. Then it may be easily shown that $\alpha F+\beta G$ is an antiderivative for $\alpha f+\beta g$.

Thus, \begin{eqnarray*} \int^b_a [\alpha f(x)+\beta g(x)]\, dx&=&(\alpha F(x)+\beta G(x))|^b_a\\ &=&(\alpha F(b)+\beta G(b))-(\alpha F(a)+\beta G(a))\\ &=&\alpha (F(b)-F(a))+\beta (G(b)-G(a))\\ &=&\alpha \int^b_a f(x)\,dx+\beta \int^b_a g(x)\, dx, \end{eqnarray*} where the first and last equalities follow from the Fundamental Theorem of Calculus.

Key Concepts

Fundamental Theorem of Calculus

Let $f$ be continuous on $[a,b]$. If $F$ is any antiderivative for $f$ on $[a,b]$, then \[\int^b_a f(t)\, dt = F(b)-F(a){\small\textrm{.}}\]

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