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Mathematics LibreTexts

7.5: The Fundamental Theorem of Calculus

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    22682
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    Theorem \(\PageIndex{1}\)

    (Fundamental Theorem of Calculus)

    Suppose \(f\) is integrable on \([a, b] .\) If \(F\) is continuous on \([a, b]\) and differentiable on \((a, b)\) with \(F^{\prime}(x)=f(x)\) for all \(x \in(a, b),\) then

    \[\int_{a}^{b} f=F(b)-F(a).\]

    Proof

    Given \(\epsilon>0,\) let \(P=\left\{x_{0}, x_{1}, \ldots, x_{n}\right\}\) be a partition of \([a, b]\) for which

    \[U(f, P)-L(f, P)<\epsilon .\]

    For \(i=1,2, \ldots, n,\) let \(t_{i} \in\left(x_{i-1}, x_{i}\right)\) be points for which

    \[F\left(x_{i}\right)-F\left(x_{i-1}\right)=f\left(t_{i}\right)\left(x_{i}-x_{i-1}\right).\]

    Then

    \[\sum_{i=1}^{n} f\left(t_{i}\right)\left(x_{i}-x_{i-1}\right)=\sum_{i=1}^{n}\left(F\left(x_{i}\right)-F\left(x_{i-1}\right)\right)=F(b)-F(a).\]

    But

    \[L(f, P) \leq \sum_{i=1}^{n} f\left(t_{i}\right)\left(x_{i}-x_{i-1}\right) \leq U(f, P),\]

    so

    \[\left|F(b)-F(a)-\int_{a}^{b} f\right|<\epsilon .\]

    Since \(\epsilon\) was arbitrary, we conclude that

    \[\int_{a}^{b} f=F(b)-F(a).\]

    Q.E.D.

    Proposition \(\PageIndex{2}\)

    (Integration by parts)

    Suppose \(f\) and \(g\) are integrable

    on \([a, b] .\) If \(F\) and \(G\) are continuous on \([a, b]\) and differentiable on \((a, b)\) with \(F^{\prime}(x)=f(x)\) and \(G^{\prime}(x)=g(x)\) for all \(x \in(a, b),\) then

    \[\int_{a}^{b} F(x) g(x) d x=F(b) G(b)-F(a) G(a)-\int_{a}^{b} f(x) G(x) d x.\]

    Proof

    By the Fundamental Theorem of Calculus,

    \[\int_{a}^{b}(F(x) g(x)+f(x) G(x)) d x=F(b) G(b)-F(a) G(a).\]

    Q.E.D.

    7.5.1 The other Fundamental Theorem of Calculus

    Proposition \(\PageIndex{3}\)

    Suppose \(f\) is integrable on \([a, b]\) and \(F:[a, b] \rightarrow \mathbb{R}\) is defined by

    \[F(x)=\int_{a}^{x} f(t) d t.\]

    Then \(F\) is uniformly continuous on \([a, b] .\)

    Proof

    Let \(\epsilon>0\) be given and let \(M>0\) be such that \(|f(x)| \leq M\) for all \(x \in[a, b] .\) Then for any \(x, y \in[a, b]\) with \(x<y\) and \(y-x<\frac{e}{M}\),

    \[|F(y)-F(x)|=\left|\int_{x}^{y} f(t) d t\right| \leq M(y-x)<\epsilon .\]

    Hence \(F\) is uniformly continuous on \([a, b] . \quad\) Q.E.D.

    The following theorem is often considered to be part of the Fundamental Theorem of Calculus.

    Theorem \(\PageIndex{4}\)

    Suppose \(f\) is integrable on \([a, b]\) and continuous at \(u \in(a, b) .\) If \(F:[a, b] \rightarrow \mathbb{R}\) is defined by

    \[F(x)=\int_{a}^{x} f(t) d t,\]

    then \(F\) is differentiable at \(u\) and \(F^{\prime}(u)=f(u)\).

    Proof

    Let \(\epsilon>0\) be given and choose \(\delta>0\) such that \(|f(x)-f(u)|<\epsilon\) whenever \(|x-u|<\delta .\) Then if \(0<h<\delta,\) we have

    \[\begin{aligned}\left|\frac{F(u+h)-F(u)}{h}-f(u)\right| &=\left|\frac{1}{h} \int_{u}^{u+h} f(t) d t-f(u)\right| \\ &=\left|\frac{1}{h} \int_{u}^{u+h}(f(t)-f(u)) d t\right| \\ &<\epsilon . \end{aligned}\]

    If \(-\delta<h<0,\) then

    \[\begin{aligned}\left|\frac{F(u+h)-F(u)}{h}-f(u)\right| &=\left|-\frac{1}{h} \int_{u+h}^{u} f(t) d t-f(u)\right| \\ &=\left|\frac{1}{h} \int_{u+h}^{u} f(t) d t+f(u)\right| \\ &=\left|\frac{1}{h} \int_{u+h}^{u} f(t) d t-\frac{1}{h} \int_{u+h}^{u} f(u) d t\right| \\ &=\left|\frac{1}{h} \int_{u+h}^{u}(f(t)-f(u)) d t\right| \\ &<\epsilon . \end{aligned}\]

    Hence

    \[F^{\prime}(u)=\lim _{h \rightarrow 0} \frac{F(u+h)-F(u)}{h}=f(u) .\]

    Q.E.D.

    Proposition \(\PageIndex{5}\)

    If \(a<b\) and \(f\) is continuous on \([a, b],\) then there exists a function \(F:[a, b] \rightarrow \mathbb{R}\) which is continuous on \([a, b]\) with \(F^{\prime}(x)=f(x)\) for all \(x \in(a, b) .\)

    Proof

    Let

    \[F(x)=\int_{a}^{x} f(t) d t .\]

    Q.E.D.

    Example \(\PageIndex{1}\)

    If

    \[g(x)=\int_{0}^{x} \sqrt{1+t^{4}} d t,\]

    then \(g^{\prime}(x)=\sqrt{1+x^{4}}\).

    Proposition \(\PageIndex{6}\)

    (Integration by substitution)

    Suppose \(I\) is an open interval, \(\varphi: I \rightarrow \mathbb{R}, a<b,[a, b] \subset I,\) and \(\varphi^{\prime}\) is continuous on \([a, b] .\) If \(f: \varphi([a, b]) \rightarrow \mathbb{R}\) is continuous, then

    \[\int_{\varphi(a)}^{\varphi(b)} f(u) d u=\int_{a}^{b} f(\varphi(x)) \varphi^{\prime}(x) d x.\]

    Proof

    If \(m\) and \(M\) are the minimum and maximum values, respectively, of \(\varphi\) on \([a, b],\) then \(\varphi([a, b])=[m, M] .\) If \(m=M,\) then \(\varphi(x)=m\) for all \(x \in[a, b],\) and both sides of \((7.5 .17)\) are \(0 .\) So we may assume \(m<M .\) Let \(F\) be a function which is continuous on \([m, M]\) with \(F^{\prime}(u)=f(u)\) for every \(u \in(m, M) .\) Let \(g=F \circ \varphi .\) Then

    \[g^{\prime}(x)=F^{\prime}(\varphi(x)) \varphi^{\prime}(x)=f(\varphi(x)) \varphi^{\prime}(x).\]

    So if \(\varphi(a) \leq \varphi(b)\),

    \[\begin{aligned} \int_{a}^{b} f(\varphi(x)) \varphi^{\prime}(x) d x &=g(b)-g(a) \\ &=F(\varphi(b))-F(\varphi(a)) \\ &=\int_{\varphi(a)}^{\varphi(b)} f(u) d u. \end{aligned}\]

    If \(\varphi(a)>\varphi(b),\) then

    \[\begin{aligned} \int_{a}^{b} f(\varphi(x)) \varphi^{\prime}(x) d x &=g(b)-g(a) \\ &=F(\varphi(b))-F(\varphi(a)) \\ &=-(F(\varphi(a))-F(\varphi(b))) \\ &=-\int_{\varphi(b)}^{\varphi(a)} f(u) d u \\ &=\int_{\varphi(a)}^{\varphi(b)} f(u) d u. \end{aligned}\]

    Q.E.D.

    Exercise \(\PageIndex{1}\)

    Evaluate

    \[\int_{0}^{1} u \sqrt{u+1} d u\]

    using (a) integration by parts and (b) substitution.

    Exercise \(\PageIndex{2}\)

    Suppose \(\varphi: \mathbb{R} \rightarrow \mathbb{R}\) is differentiable on \(\mathbb{R}\) and periodic with period \(1(\text { that is, } \varphi(x+1)=\varphi(x) \text { for every } x \in \mathbb{R}) .\) Show that for any continuous function \(f: \mathbb{R} \rightarrow \mathbb{R}\),

    \[\int_{0}^{1} f(\varphi(x)) \varphi^{\prime}(x) d x=0.\]

    Theorem \(\PageIndex{7}\)

    (Integral Mean Value Theorem)

    If \(f\) is continuous on \([a, b],\) then there exists \(c \in[a, b]\) such that

    \[\int_{a}^{b} f=f(c)(b-a).\]

    Exercise \(\PageIndex{3}\)

    Prove the Integral Mean Value Theorem.

    Theorem \(\PageIndex{8}\)

    (Generalized Integral Mean Value Theorem)

    If \(f\) and \(g\) are continuous on \([a, b]\) and \(g(x)>0\) for all \(x \in[a, b],\) then there exists \(c \in[a, b]\) such that

    \[\int_{a}^{b} f g=f(c) \int_{a}^{b} g.\]

    Exercise \(\PageIndex{4}\)

    Prove the Generalized Integral Mean Value Theorem.

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