MATH 409 Lecture 20

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Lecture Slides

Exam next Thursday

Homework problems just require definition of improper Riemann definition; look at hint for problem 5.3.12

Integral with Variable Limit

(See Integral with a variable limit→)

Theorem. If $f$ is continuous at a point $x\in [a,b]$, then $F$ is differentiable at $x$ and $F'(x)=f(x)$

Proof. For any $x,y\in [a,b]$ such that $x<y$, we have

$\int _{a}^{y}f(t)\,\mathrm {d} t=\int _{a}^{x}f(t)\,\mathrm {d} t+\int _{x}^{y}f(t)\,\mathrm {d} t\,\!$

Then

$F(y)-F(x)-f(x)(y-x)=\int _{x}^{y}f(t)\,\mathrm {d} t-\int _{x}^{y}f(x)\,\mathrm {d} t$

So that

$\left|F(y)-F(x)-f(x)(y-x)\right|=\left|\int _{x}^{y}\left(f(t)-f(x)\right)\,\mathrm {d} t\right|\leq \int _{x}^{y}\left|f(t)-f(x)\right|\,\mathrm {d} t\leq \sup _{t\in [x,y]}\left|f(t)-f(x)\right|\,(y-x)$

Dividing both sides by Failed to parse (Conversion error. Server ("https://wikimedia.org/api/rest_") reported: "Cannot get mml. Server problem."): {\displaystyle (y-x)} gives

${\frac {F(y)-F(x)}{y-x}}-f(x)\leq \sup _{t\in [x,y]}\left|f(t)-f(x)\right|$

If the function $f$ is right continuous at $x$ (i.e. $f(y)\to f(x)$ as $y\to x^{+}$), then $\sup _{t\in [x,y]}\left|f(t)-f(x)\right|\to 0$ as $y\to x^{+}$. It follows that $f(x)$ is the right-hand derivative of $F$ at $x$. Likewise, one can prove that left continuity of f at x implies $f(x)$ is the left derivative at $x$.

quod erat demonstrandum

The Fundamental Theorem of Calculus

Theorem. [Part I]. If a function $f$ is continuous on an interval $[a,b]$, then the function

$F(x)=\int _{a}^{x}f(t)\,\mathrm {d} t\qquad x\in [a,b]$

is continuously differentiable on $[a,b]$. Moreover, $F'(x)=f(x)$ for all $x\in [a,b]$.

Proof. Since $f$ is continuous, it is also integrable on $[a,b]$. As proved earlier, integrability of Failed to parse (MathML with SVG or PNG fallback (recommended for modern browsers and accessibility tools): Invalid response ("Math extension cannot connect to Restbase.") from server "https://wikimedia.org/api/rest_v1/":): {\displaystyle f} implies that the function Failed to parse (MathML with SVG or PNG fallback (recommended for modern browsers and accessibility tools): Invalid response ("Math extension cannot connect to Restbase.") from server "https://wikimedia.org/api/rest_v1/":): {\displaystyle F} is well-defined on Failed to parse (MathML with SVG or PNG fallback (recommended for modern browsers and accessibility tools): Invalid response ("Math extension cannot connect to Restbase.") from server "https://wikimedia.org/api/rest_v1/":): {\displaystyle [a,b]} . Moreover, Failed to parse (MathML with SVG or PNG fallback (recommended for modern browsers and accessibility tools): Invalid response ("Math extension cannot connect to Restbase.") from server "https://wikimedia.org/api/rest_v1/":): {\displaystyle F'(x) = f(x)} whenever Failed to parse (MathML with SVG or PNG fallback (recommended for modern browsers and accessibility tools): Invalid response ("Math extension cannot connect to Restbase.") from server "https://wikimedia.org/api/rest_v1/":): {\displaystyle f} is continuous at the point Failed to parse (MathML with SVG or PNG fallback (recommended for modern browsers and accessibility tools): Invalid response ("Math extension cannot connect to Restbase.") from server "https://wikimedia.org/api/rest_v1/":): {\displaystyle x} . Therefore, the continuoity of Failed to parse (MathML with SVG or PNG fallback (recommended for modern browsers and accessibility tools): Invalid response ("Math extension cannot connect to Restbase.") from server "https://wikimedia.org/api/rest_v1/":): {\displaystyle f} on $[a,b]$ implies that $F'(x)=f(x)$ for all $x\in [a,b]$. In particular, $F$ is continuously differentiable on $[a,b]$.

quod erat demonstrandum

Theorem. [Part II]. If a function $F$ is differentiable on $[a,b]$ and the derivative $F'$ is integrable on $[a,b]$, then

$\int _{a}^{x}F'(t)\,\mathrm {d} t=F(x)-F(a)$ for all $x\in [a,b]$

Proof. The case $x=a$ is trivial: $0=0$. Assume $x\in (a,b]$. Since $F'$ is integrable on $[a,b]$, it is also integrable on any subinterval $[a,x]$, $x\in (a,b)$. Therefore, it is no loss to assume that $x=b$.

Consider an arbitrary partition $P=\left\{x_{0},x_{1},\ldots ,x_{n}\right\}$ of $[a,b]$. Let us choose samples $t_{j}\in [x_{j-1},x_{j}]$ for the Riemann sum ${\mathcal {S}}(F',P,t_{j})$ so that $F(x_{j})-F(x_{j-1})=F'(t_{j})\,(x_{j}-x_{j-1})$ (this is possible due to the Mean Value Theorem). Then

${\begin{aligned}{\mathcal {S}}(F',P,t_{j})&=\sum _{j=1}^{n}F'(t_{j})\,(x_{j}-x_{j-1})\\&=\sum _{j=1}^{n}\left(F(x_{j})-F(x_{j-1})\right)\\&=F(x_{n})-F(x_{0})\\&=F(b)-F(a)\end{aligned}}$

Since the sums ${\mathcal {S}}(F',P,t_{j})$ converge to $\int _{a}^{b}F'(t)\,\mathrm {d} t$ as $\left\|P\right\|\to 0$, the theorem follows: Failed to parse (MathML with SVG or PNG fallback (recommended for modern browsers and accessibility tools): Invalid response ("Math extension cannot connect to Restbase.") from server "https://wikimedia.org/api/rest_v1/":): {\displaystyle \int_a^b F'(t) \, \mathrm{d}t = F(b) - F(a)} .

quod erat demonstrandum

The Indefinite Integral

Given a function Failed to parse (MathML with SVG or PNG fallback (recommended for modern browsers and accessibility tools): Invalid response ("Math extension cannot connect to Restbase.") from server "https://wikimedia.org/api/rest_v1/":): {\displaystyle f : [a,b] \to \mathbb{R}} , a function $F:[a,b]\to \mathbb {R}$ is called the indefinite integral (or antiderivative, primitive integral, or the primitive) of $f$ if $F'(x)=f(x)$ for all $x\in [a,b]$. Notation:

$\int f(x)\,\mathrm {d} x$

If the function $f$ is continuous on $[a,b]$, then the function $F(x)=\int _{a}^{x}f(t)\,\mathrm {d} t$ for $x\in [a,b]$ is an indefinite integral of $f$ due to the Fundamental Theorem of Calculus.

Suppose $F$ is an antiderivative of $f$. If $G$ is another antiderivative of $f$, then Failed to parse (MathML with SVG or PNG fallback (recommended for modern browsers and accessibility tools): Invalid response ("Math extension cannot connect to Restbase.") from server "https://wikimedia.org/api/rest_v1/":): {\displaystyle G' = F'} on Failed to parse (MathML with SVG or PNG fallback (recommended for modern browsers and accessibility tools): Invalid response ("Math extension cannot connect to Restbase.") from server "https://wikimedia.org/api/rest_v1/":): {\displaystyle [a,b]} . Hence Failed to parse (MathML with SVG or PNG fallback (recommended for modern browsers and accessibility tools): Invalid response ("Math extension cannot connect to Restbase.") from server "https://wikimedia.org/api/rest_v1/":): {\displaystyle (G-F)' = G' - F' = 0} on Failed to parse (MathML with SVG or PNG fallback (recommended for modern browsers and accessibility tools): Invalid response ("Math extension cannot connect to Restbase.") from server "https://wikimedia.org/api/rest_v1/":): {\displaystyle [a,b]} . It follows that $G-F$ is a constant function. Conversely, for any constant Failed to parse (MathML with SVG or PNG fallback (recommended for modern browsers and accessibility tools): Invalid response ("Math extension cannot connect to Restbase.") from server "https://wikimedia.org/api/rest_v1/":): {\displaystyle C} , the function Failed to parse (MathML with SVG or PNG fallback (recommended for modern browsers and accessibility tools): Invalid response ("Math extension cannot connect to Restbase.") from server "https://wikimedia.org/api/rest_v1/":): {\displaystyle G(x) = F(x) + C} is also an antiderivative of Failed to parse (MathML with SVG or PNG fallback (recommended for modern browsers and accessibility tools): Invalid response ("Math extension cannot connect to Restbase.") from server "https://wikimedia.org/api/rest_v1/":): {\displaystyle f} .

$\int f(x)\,\mathrm {d} x=F(x)+C$

Where $C$ is an arbitrary constant.

Examples

• Failed to parse (MathML with SVG or PNG fallback (recommended for modern browsers and accessibility tools): Invalid response ("Math extension cannot connect to Restbase.") from server "https://wikimedia.org/api/rest_v1/":): {\displaystyle \int x^\alpha \,\mathrm{d}x = \frac{x^{\alpha+1}}{\alpha+1} + C} on Failed to parse (MathML with SVG or PNG fallback (recommended for modern browsers and accessibility tools): Invalid response ("Math extension cannot connect to Restbase.") from server "https://wikimedia.org/api/rest_v1/":): {\displaystyle (0, \infty)} for Failed to parse (MathML with SVG or PNG fallback (recommended for modern browsers and accessibility tools): Invalid response ("Math extension cannot connect to Restbase.") from server "https://wikimedia.org/api/rest_v1/":): {\displaystyle \alpha \ne 1} .
• Failed to parse (MathML with SVG or PNG fallback (recommended for modern browsers and accessibility tools): Invalid response ("Math extension cannot connect to Restbase.") from server "https://wikimedia.org/api/rest_v1/":): {\displaystyle \int \frac{1}{x} \,\mathrm{d}x = \log{x} + C} on $(0,\infty )$.
• Failed to parse (MathML with SVG or PNG fallback (recommended for modern browsers and accessibility tools): Invalid response ("Math extension cannot connect to Restbase.") from server "https://wikimedia.org/api/rest_v1/":): {\displaystyle \int \sin{x} = -\cos{x} + C}
• Failed to parse (MathML with SVG or PNG fallback (recommended for modern browsers and accessibility tools): Invalid response ("Math extension cannot connect to Restbase.") from server "https://wikimedia.org/api/rest_v1/":): {\displaystyle \int \cos{x} = \sin{x} + C}

Integration by Parts

Theorem. Suppose that functions Failed to parse (MathML with SVG or PNG fallback (recommended for modern browsers and accessibility tools): Invalid response ("Math extension cannot connect to Restbase.") from server "https://wikimedia.org/api/rest_v1/":): {\displaystyle f} and $g$ are differentiable on $[a,b]$ with derivatives Failed to parse (MathML with SVG or PNG fallback (recommended for modern browsers and accessibility tools): Invalid response ("Math extension cannot connect to Restbase.") from server "https://wikimedia.org/api/rest_v1/":): {\displaystyle f'} and Failed to parse (MathML with SVG or PNG fallback (recommended for modern browsers and accessibility tools): Invalid response ("Math extension cannot connect to Restbase.") from server "https://wikimedia.org/api/rest_v1/":): {\displaystyle g'} integrable on Failed to parse (MathML with SVG or PNG fallback (recommended for modern browsers and accessibility tools): Invalid response ("Math extension cannot connect to Restbase.") from server "https://wikimedia.org/api/rest_v1/":): {\displaystyle [a,b]} . Then

Failed to parse (MathML with SVG or PNG fallback (recommended for modern browsers and accessibility tools): Invalid response ("Math extension cannot connect to Restbase.") from server "https://wikimedia.org/api/rest_v1/":): {\displaystyle \int_{a}^{b} f(x) \, g'(x) \,\mathrm{d}x = f(b) \, g(b) - f(a) \, g(a) - \int_{a}^{b} f'(x) \, g(x) \,\mathrm{d}x}

Proof. By the product rule, $(f\,g)'=f'\,g+f\,g'$ on $[a,b]$. Since Failed to parse (MathML with SVG or PNG fallback (recommended for modern browsers and accessibility tools): Invalid response ("Math extension cannot connect to Restbase.") from server "https://wikimedia.org/api/rest_v1/":): {\displaystyle f, f', g, g'} are integrable on Failed to parse (MathML with SVG or PNG fallback (recommended for modern browsers and accessibility tools): Invalid response ("Math extension cannot connect to Restbase.") from server "https://wikimedia.org/api/rest_v1/":): {\displaystyle [a,b]} by the hypothesis, so are the products Failed to parse (MathML with SVG or PNG fallback (recommended for modern browsers and accessibility tools): Invalid response ("Math extension cannot connect to Restbase.") from server "https://wikimedia.org/api/rest_v1/":): {\displaystyle f' \, g} and Failed to parse (MathML with SVG or PNG fallback (recommended for modern browsers and accessibility tools): Invalid response ("Math extension cannot connect to Restbase.") from server "https://wikimedia.org/api/rest_v1/":): {\displaystyle f \, g'} . Then Failed to parse (MathML with SVG or PNG fallback (recommended for modern browsers and accessibility tools): Invalid response ("Math extension cannot connect to Restbase.") from server "https://wikimedia.org/api/rest_v1/":): {\displaystyle (f\,g)'} is integrable on Failed to parse (MathML with SVG or PNG fallback (recommended for modern browsers and accessibility tools): Invalid response ("Math extension cannot connect to Restbase.") from server "https://wikimedia.org/api/rest_v1/":): {\displaystyle [a,b]} as well. By the Fundamental Theorem of Calculus,

Failed to parse (MathML with SVG or PNG fallback (recommended for modern browsers and accessibility tools): Invalid response ("Math extension cannot connect to Restbase.") from server "https://wikimedia.org/api/rest_v1/":): {\displaystyle \begin{align} f(b) \, g(b) - f(a) \, g(a) &= \int_{a}^{b} (f\,g)'(x) \,\mathrm{d}x \\ &= \int_{a}^{b} f'(x) \, g(x) \,\mathrm{d}x + \int_{a}^{b} f(x) \, g'(x) \,\mathrm{d}x \end{align}}

quod erat demonstrandum

Corollary. Suppose that functions Failed to parse (MathML with SVG or PNG fallback (recommended for modern browsers and accessibility tools): Invalid response ("Math extension cannot connect to Restbase.") from server "https://wikimedia.org/api/rest_v1/":): {\displaystyle f,g} are continuously differentiable on Failed to parse (MathML with SVG or PNG fallback (recommended for modern browsers and accessibility tools): Invalid response ("Math extension cannot connect to Restbase.") from server "https://wikimedia.org/api/rest_v1/":): {\displaystyle [a,b]} . Then

Failed to parse (MathML with SVG or PNG fallback (recommended for modern browsers and accessibility tools): Invalid response ("Math extension cannot connect to Restbase.") from server "https://wikimedia.org/api/rest_v1/":): {\displaystyle \int f(x) \, g'(x) \,\mathrm{d}x = f(x) \, g(x) - \int f'(x) \, g(x) \,\mathrm{d}x} on Failed to parse (MathML with SVG or PNG fallback (recommended for modern browsers and accessibility tools): Invalid response ("Math extension cannot connect to Restbase.") from server "https://wikimedia.org/api/rest_v1/":): {\displaystyle [a,b]}

To simplify notation, it is convenient to use the Leibniz differential $\mathrm {d} f$ of a function Failed to parse (MathML with SVG or PNG fallback (recommended for modern browsers and accessibility tools): Invalid response ("Math extension cannot connect to Restbase.") from server "https://wikimedia.org/api/rest_v1/":): {\displaystyle f} defined by Failed to parse (MathML with SVG or PNG fallback (recommended for modern browsers and accessibility tools): Invalid response ("Math extension cannot connect to Restbase.") from server "https://wikimedia.org/api/rest_v1/":): {\displaystyle \mathrm{d}f(x) = f'(x) \,\mathrm{d}x = \frac{\mathrm{d}f}{\mathrm{d}x}\,\mathrm{d}x} . Another convenient notation is $\left.f(x)\right|_{x=a}^{b}$ or simply Failed to parse (MathML with SVG or PNG fallback (recommended for modern browsers and accessibility tools): Invalid response ("Math extension cannot connect to Restbase.") from server "https://wikimedia.org/api/rest_v1/":): {\displaystyle \left. f(x) \right|_a^b} , which denotes the difference Failed to parse (MathML with SVG or PNG fallback (recommended for modern browsers and accessibility tools): Invalid response ("Math extension cannot connect to Restbase.") from server "https://wikimedia.org/api/rest_v1/":): {\displaystyle f(b) - f(a)} .

Now the formula of integration by parts can be rewritten as

$\int _{a}^{b}f(x)\,\mathrm {d} g(x)=\left.f(x)\,g(x)\right|_{a}^{b}-\int _{a}^{b}g(x)\,\mathrm {d} f(x)$

for definite integrals, and as

$\int f\,\mathrm {d} g=f\,g-\int g\,\mathrm {d} f$

for indefinite integrals

Examples

• Failed to parse (MathML with SVG or PNG fallback (recommended for modern browsers and accessibility tools): Invalid response ("Math extension cannot connect to Restbase.") from server "https://wikimedia.org/api/rest_v1/":): {\displaystyle \int \log{x} \,\mathrm{d}x = x \, \log{x} - \int x \,\mathrm{d}(\log{x}) = x \, \log{x} - \int \mathrm{d}x = x \, \log{x} - x + C}
• Failed to parse (MathML with SVG or PNG fallback (recommended for modern browsers and accessibility tools): Invalid response ("Math extension cannot connect to Restbase.") from server "https://wikimedia.org/api/rest_v1/":): {\displaystyle \int_0^{\frac{\pi}{2}} x \, \sin{x} \,\mathrm{d}x = \left. -x \, \cos{x} \right|_0^{\frac{\pi}{2}} - \int_0^{\frac{\pi}{2}} (-\cos{x}) \,\mathrm{d}x = \int_0^{\frac{\pi}{2}} \cos{x} \,\mathrm{d}x = \left. \sin{x} \right|_0^{\frac{\pi}{2}} = 1} .

Change of the variable in an integral

(commonly called u-substitution as Failed to parse (MathML with SVG or PNG fallback (recommended for modern browsers and accessibility tools): Invalid response ("Math extension cannot connect to Restbase.") from server "https://wikimedia.org/api/rest_v1/":): {\displaystyle \phi} is often replaced by Failed to parse (MathML with SVG or PNG fallback (recommended for modern browsers and accessibility tools): Invalid response ("Math extension cannot connect to Restbase.") from server "https://wikimedia.org/api/rest_v1/":): {\displaystyle u} in practice.)

Theorem. if Failed to parse (MathML with SVG or PNG fallback (recommended for modern browsers and accessibility tools): Invalid response ("Math extension cannot connect to Restbase.") from server "https://wikimedia.org/api/rest_v1/":): {\displaystyle \phi} is continuously differentiable on a closed, nondegenerate interval Failed to parse (MathML with SVG or PNG fallback (recommended for modern browsers and accessibility tools): Invalid response ("Math extension cannot connect to Restbase.") from server "https://wikimedia.org/api/rest_v1/":): {\displaystyle [a,b]} and Failed to parse (MathML with SVG or PNG fallback (recommended for modern browsers and accessibility tools): Invalid response ("Math extension cannot connect to Restbase.") from server "https://wikimedia.org/api/rest_v1/":): {\displaystyle f} is continuous on Failed to parse (MathML with SVG or PNG fallback (recommended for modern browsers and accessibility tools): Invalid response ("Math extension cannot connect to Restbase.") from server "https://wikimedia.org/api/rest_v1/":): {\displaystyle \phi([a,b])} , then

$\int _{\phi (a)}^{\phi (b)}f(t)\,\mathrm {d} t=\int _{a}^{b}f(\phi (x))\,\phi '(x)\,\mathrm {d} x=\int _{a}^{b}f(\phi (x))\,\mathrm {d} \phi (x)$

Be aware that Failed to parse (MathML with SVG or PNG fallback (recommended for modern browsers and accessibility tools): Invalid response ("Math extension cannot connect to Restbase.") from server "https://wikimedia.org/api/rest_v1/":): {\displaystyle t = \phi(x)} is a proper change of variable only if the function Failed to parse (MathML with SVG or PNG fallback (recommended for modern browsers and accessibility tools): Invalid response ("Math extension cannot connect to Restbase.") from server "https://wikimedia.org/api/rest_v1/":): {\displaystyle \phi} is strictly monotone. However, the theorem holds even without this assumption.

Proof. Let us define two functions

Failed to parse (MathML with SVG or PNG fallback (recommended for modern browsers and accessibility tools): Invalid response ("Math extension cannot connect to Restbase.") from server "https://wikimedia.org/api/rest_v1/":): {\displaystyle \begin{align} F(u) &= \int_{\phi(a)}^{u} f(t) \,\mathrm{d}t \qquad u \in \phi([a,b]) \\ G(u) &= \int_{a}^{x} f(\phi(s)) \, \phi'(s) \,\mathrm{d}s \qquad x \in [a,b] \end{align}}

It follows from the Fundamental Theorem of Calculus that Failed to parse (MathML with SVG or PNG fallback (recommended for modern browsers and accessibility tools): Invalid response ("Math extension cannot connect to Restbase.") from server "https://wikimedia.org/api/rest_v1/":): {\displaystyle F'(u) = f(u)} and $G'(x)=f(\phi (x))\,\phi '(x)$. By the Chain Rule, Failed to parse (MathML with SVG or PNG fallback (recommended for modern browsers and accessibility tools): Invalid response ("Math extension cannot connect to Restbase.") from server "https://wikimedia.org/api/rest_v1/":): {\displaystyle (F \circ \phi)'(x) = F'(\phi(x)) \, \phi'(x) = f(\phi(x)) \, \phi'(x) = G'(x)} .

Therefore, Failed to parse (MathML with SVG or PNG fallback (recommended for modern browsers and accessibility tools): Invalid response ("Math extension cannot connect to Restbase.") from server "https://wikimedia.org/api/rest_v1/":): {\displaystyle \left( F(\phi(x)) - G(x) \right)' = 0} for all Failed to parse (MathML with SVG or PNG fallback (recommended for modern browsers and accessibility tools): Invalid response ("Math extension cannot connect to Restbase.") from server "https://wikimedia.org/api/rest_v1/":): {\displaystyle x \in [a,b]} . It follows that the function Failed to parse (MathML with SVG or PNG fallback (recommended for modern browsers and accessibility tools): Invalid response ("Math extension cannot connect to Restbase.") from server "https://wikimedia.org/api/rest_v1/":): {\displaystyle F(\phi(x)) - G(x)} is constant on Failed to parse (MathML with SVG or PNG fallback (recommended for modern browsers and accessibility tools): Invalid response ("Math extension cannot connect to Restbase.") from server "https://wikimedia.org/api/rest_v1/":): {\displaystyle [a,b]} . In particular, Failed to parse (MathML with SVG or PNG fallback (recommended for modern browsers and accessibility tools): Invalid response ("Math extension cannot connect to Restbase.") from server "https://wikimedia.org/api/rest_v1/":): {\displaystyle F(\phi(b)) - G(b) = F(\phi(a)) - G(a) = 0 - 0 = 0} . Hence Failed to parse (MathML with SVG or PNG fallback (recommended for modern browsers and accessibility tools): Invalid response ("Math extension cannot connect to Restbase.") from server "https://wikimedia.org/api/rest_v1/":): {\displaystyle F(\phi(b)) = G(b)} .

quod erat demonstrandum

Note: It is possible that Failed to parse (MathML with SVG or PNG fallback (recommended for modern browsers and accessibility tools): Invalid response ("Math extension cannot connect to Restbase.") from server "https://wikimedia.org/api/rest_v1/":): {\displaystyle \phi(a) \ge \phi(b)} . To make sense of this case, we set Failed to parse (MathML with SVG or PNG fallback (recommended for modern browsers and accessibility tools): Invalid response ("Math extension cannot connect to Restbase.") from server "https://wikimedia.org/api/rest_v1/":): {\displaystyle \int_{c}^{d} f(t) \,\mathrm{d}t {{=}} -\int_{d}^{c} f(t) \,\mathrm{d}t} if Failed to parse (MathML with SVG or PNG fallback (recommended for modern browsers and accessibility tools): Invalid response ("Math extension cannot connect to Restbase.") from server "https://wikimedia.org/api/rest_v1/":): {\displaystyle c > d} . Also we set the integral to be 0 if Failed to parse (MathML with SVG or PNG fallback (recommended for modern browsers and accessibility tools): Invalid response ("Math extension cannot connect to Restbase.") from server "https://wikimedia.org/api/rest_v1/":): {\displaystyle c = d} .