MA005: Calculus I

If we know and can evaluate some antiderivative of a function, then we can evaluate any definite integral of that function.

The Fundamental Theorem of Calculus (Part 2)

If $\mathrm{f}(x)$ is continuous and $\mathrm{F}(x)$ is any antiderivative of $\mathrm{f}\left(\mathrm{F}^{\prime}(x)=\mathrm{f}(x)\right)$,

then $\int_{\mathrm{a}}^{\mathrm{b}} \mathrm{f}(x) \mathrm{d} \mathrm{x}=\left.\mathrm{F}(x)\right|_{\mathrm{a}} ^{\mathrm{b}}=\mathrm{F}(\mathrm{b})-\mathrm{F}(\mathrm{a})$.

Proof: If $F$ is an antiderivative of $f$, then $\mathrm{A}(x)=\int_{\mathrm{a}}^{x} \mathrm{f}(t) \mathrm{dt}$ are both antiderivatives of $f$, $\mathrm{F}^{\prime}(x)=\mathrm{f}(x)=\mathrm{A}^{\prime}(x)$, so $F$ and $A$ differ by a constant: $\mathrm{A}(x)-\mathrm{F}(x)=\mathrm{C}$ for all $x$.

At $x=\mathrm{a}$, we have $\mathrm{C}=\mathrm{A}(\mathrm{a})-\mathrm{F}(\mathrm{a})=0-\mathrm{F}(\mathrm{a})=-\mathrm{F}(\mathrm{a}) \text { so } \mathrm{C}=-\mathrm{F}(\mathrm{a})$ and the equation $\mathrm{A}(x)-\mathrm{F}(x)=\mathrm{C}$ becomes $\mathrm{A}(x)-\mathrm{F}(x)=-\mathrm{F}(\mathrm{a})$. Then $\mathrm{A}(x)=\mathrm{F}(x)-\mathrm{F}(\mathrm{a})$ for all $x$ so $\mathrm{A}(\mathrm{b})=\mathrm{F}(\mathrm{b})-\mathrm{F}(\mathrm{a})$ and $\int_{\mathrm{a}}^{\mathbf{b}} \mathrm{f}(x) \mathrm{d} \mathrm{x}=\mathrm{A}(\mathbf{b})=\mathrm{F}(\mathbf{b})-\mathrm{F}(\mathrm{a})$, the formula we wanted.

The definite integral of a continuous function $f$ can be found by finding an antiderivative of $f$ (any antiderivative of $f$ will work) and then doing some arithmetic with this antiderivative. The theorem does not tell us how to find an antiderivative of $f$, and it does not tell us how to find the definite integral of a discontinuous function. It is possible to evaluate definite integrals of some discontinuous functions (Section 4.3), but the Fundamental Theorem of Calculus can not be used to do so.

Example 3: Evaluate $\int_{0}^{2}\left(x^{2}-1\right) \mathrm{dx}$.

Solution: $\mathrm{F}(\mathrm{x})=\frac{x^{3}}{3}-x$ is an antiderivative of $\mathrm{f}(x)=x^{2}-1$ (check that $\mathrm{D}\left(\frac{x^{3}}{3}-x\right)=x^{2}-1$ ) so

$\int_{0}^{2}\left(x^{2}-1\right) \mathrm{dx}=\frac{x^{3}}{3}-\left.x\right|_{0} ^{2}=\left\{\frac{2^{3}}{3}-2\right\}-\left\{\frac{0^{3}}{3}-0\right\}=2 / 3-0=2 / 3$.

If friends had picked a different antiderivative of $x^{2}-1$, say $\mathrm{F}(x)=\frac{x^{3}}{3}-x+4$, then their calculations would be slightly different but the result would be the same:

$\int_{0}^{2}\left(x^{2}-1\right) \mathrm{dx}=\frac{x^{3}}{3}-x+\left.4\right|_{0} ^{2}=\left(\frac{2^{3}}{3}-2+4\right)-\left(\frac{0^{3}}{3}-0+4\right)=14 / 3- 4 = 2/3$

Practice 2: Evaluate $\int_{1}^{3}\left(3 x^{2}-1\right) \mathrm{dx}$.

Example 4: Evaluate $\int_{1.5}^{2.7} \mathrm{INT}(x) \mathrm{dx}$. ($INT(x)$ is the largest integer less than or equal to $x$. Fig. 6)

Solution: $\mathrm{f}(x)=\operatorname{INT}(x)$ is not continuous at $x = 2$ in the interval [1.5, 2.7] so the Fundamental Theorem of Calculus can not be used. We can, however, use our understanding of the meaning of an integral to get

$\begin{aligned} &\int_{1.5}^{2.7} \operatorname{INT}(x) \mathrm{dx}=(\text { area for } x \text { between } 1.5 \text { and } 2)+(\text { area for } x \text { between } 2 \text { and 2.7) } \\ &=(\text { base })(\text { height })+(\text { base })(\text { height })=(.5)(1)+(.7)(2)=1.9 \end{aligned}$

Practice 3: Evaluate $\int_{1.3}^{3.4} \mathrm{INT}(x) \mathrm{dx}$.

Calculus is the study of derivatives and integrals, their meanings and their applications. The Fundamental Theorem of Calculus shows that differentiation and integration are closely related and that integration is really antidifferentiation, the inverse of differentiation.