Practice / Calculus

The fundamental theorem of calculus: derivatives of integrals and definite integrals

Ten problems on both parts of the fundamental theorem: an accumulation function checked against its closed form, a chain-rule upper limit, a variable lower limit, two moving limits and the maximum they give, critical points of an integral, definite integrals with an absolute value, the derivative of a Gaussian and an exponential CDF, and a limit of an integral by L'Hôpital's rule, with worked solutions and the mistakes of a missing chain-rule factor, a lost minus sign and confusing F with f.

Before you start

The fundamental theorem is the link between the two halves of calculus: the derivative of an accumulated area is the height of the curve, and a definite integral is a difference of antiderivatives. In machine learning it is the reason a probability density is the derivative of its cumulative distribution function, so the density of any transformed variable comes from the chain rule on a CDF. The work goes wrong in a few places: a chain-rule factor left off a limit like $x^3$ the minus sign of a variable lower limit dropped, the value of the integrand at a constant lower limit subtracted, an absolute value integrated as if it were not there, and the accumulation function $F$ confused with its integrand $f$ These ten problems cover each of them.

Part 1. If $f$ is continuous and $F(x) = \displaystyle\int_a^x f(t)\,dt$ then $F'(x) = f(x)$ The constant $a$ does not appear in the answer.
Moving limits. By the chain rule, $\dfrac{d}{dx}\displaystyle\int_{g(x)}^{h(x)} f(t)\,dt = f\big(h(x)\big)\,h'(x) - f\big(g(x)\big)\,g'(x)$ A constant limit contributes $0$ a variable lower limit contributes a minus sign.
Part 2. If $f$ is continuous on $[a, b]$ and $F' = f$ then $\displaystyle\int_a^b f(x)\,dx = F(b) - F(a)$ Any antiderivative works, so no $+C$ is needed.
$F(x) = \int_a^x f(t)\,dt$ is a function of $x$ not of $t$ $t$ is a dummy variable. $F$ is increasing where $f > 0$ so its critical points are the zeros of $f$
A cumulative distribution function is $F(x) = \displaystyle\int_{-\infty}^x p(t)\,dt$ for a density $p$ so $F' = p$ For the standard normal, $\Phi(z) = \displaystyle\int_{-\infty}^z \varphi(t)\,dt$ with $\varphi(t) = \dfrac{1}{\sqrt{2\pi}}e^{-t^2/2}$
$\lvert x - c\rvert$ is $c - x$ for $x < c$ and $x - c$ for $x > c$ split the integral at $c$

Builds on: Differentiation rules: chain, product, quotient

Problems

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Let $F(x) = \displaystyle\int_1^x (t^3 - 2t)\,dt$ Find $F'(x)$ by Part 1, then find $F(x)$ in closed form and check.
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Find $\dfrac{d}{dx}\displaystyle\int_1^{x^3} \ln t\,dt$ for $x > 0$ and its value at $x = 2$
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Find $\dfrac{d}{dx}\displaystyle\int_x^5 \sqrt{1 + t^4}\,dt$ and its value at $x = 1$
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Let $G(x) = \displaystyle\int_x^{2x} e^{-t}\,dt$ for $x \ge 0$ Find $G'(x)$ and the $x$ at which $G$ is largest.
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Let $F(x) = \displaystyle\int_0^x (t^2 - 4t + 3)\,dt$ Find the local maximum and minimum values of $F$
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Evaluate $\displaystyle\int_1^4 \Big(2x + \frac{1}{\sqrt x}\Big)\,dx$
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Evaluate $\displaystyle\int_0^3 \lvert x - 1\rvert\,dx$
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$\Phi$ is the standard normal CDF. Find $\dfrac{d}{dx}\Phi(2x - 1)$ and its value at $x = \tfrac12$
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The waiting time $T$ has density $p(t) = 2e^{-2t}$ for $t \ge 0$ Find its CDF $F(x) = P(T \le x)$ check that $F' = p$ and find the median.
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Find $\displaystyle\lim_{x\to0}\frac{1}{x^3}\int_0^x \sin(t^2)\,dt$

Answers

$F'(x) = x^3 - 2x$ and $F(x) = \tfrac{x^4}4 - x^2 + \tfrac34$ has that derivative
$9x^2\ln x$ at $x = 2$ it is $36\ln2 \approx 24.95$
$-\sqrt{1 + x^4}$ at $x = 1$ it is $-\sqrt2 \approx -1.414$
$G'(x) = 2e^{-2x} - e^{-x}$ $G$ is largest at $x = \ln2 \approx 0.693$ where $G = \tfrac14$
Local maximum $\tfrac43$ at $x = 1$ local minimum $0$ at $x = 3$
$17$
$\tfrac52$
$2\varphi(2x - 1)$ at $x = \tfrac12$ it is $\sqrt{2/\pi} \approx 0.798$
$F(x) = 1 - e^{-2x}$ median $\tfrac{\ln2}2 \approx 0.347$
$\tfrac13$

Worked solutions

Problem 1

Let $F(x) = \displaystyle\int_1^x (t^3 - 2t)\,dt$ Find $F'(x)$ by Part 1, then find $F(x)$ in closed form and check.

$F'(x) = x^3 - 2x$ Part 1: replace $t$ with the upper limit.
$F(x) = \Big[\tfrac{t^4}4 - t^2\Big]_1^x = \tfrac{x^4}4 - x^2 - \big(\tfrac14 - 1\big)$ Part 2 with the antiderivative $\tfrac{t^4}4 - t^2$
$F(x) = \tfrac{x^4}4 - x^2 + \tfrac34$ The lower limit only adds the constant $\tfrac34$
$F'(x) = x^3 - 2x$ and $F(x) = \tfrac{x^4}4 - x^2 + \tfrac34$ has that derivativeThe constant $\tfrac34$ differentiates to $0$ which is why the lower limit never appears in $F'$ Also $F(1) = 0$ as it must.

Problem 2

Find $\dfrac{d}{dx}\displaystyle\int_1^{x^3} \ln t\,dt$ for $x > 0$ and its value at $x = 2$

$\ln(x^3)\cdot\dfrac{d}{dx}x^3$ Put the upper limit into the integrand, then multiply by its derivative.
$= 3\ln x\cdot3x^2$ $\ln(x^3) = 3\ln x$ and $(x^3)' = 3x^2$
Check: $\displaystyle\int_1^{x^3}\ln t\,dt = \big[t\ln t - t\big]_1^{x^3} = 3x^3\ln x - x^3 + 1$ whose derivative is $9x^2\ln x + 3x^2 - 3x^2$ The long way agrees.
$9x^2\ln x$ at $x = 2$ it is $36\ln2 \approx 24.95$ Without the chain-rule factor the answer would be $3\ln x$ off by a factor of $3x^2$

Problem 3

Find $\dfrac{d}{dx}\displaystyle\int_x^5 \sqrt{1 + t^4}\,dt$ and its value at $x = 1$

$\displaystyle\int_x^5 \sqrt{1 + t^4}\,dt = -\int_5^x \sqrt{1 + t^4}\,dt$ Swapping the limits changes the sign, and puts $x$ in the upper limit.
$\dfrac{d}{dx}\Big(-\displaystyle\int_5^x \sqrt{1 + t^4}\,dt\Big) = -\sqrt{1 + x^4}$ Part 1.
$-\sqrt{1 + x^4}$ at $x = 1$ it is $-\sqrt2 \approx -1.414$ Moving the lower limit right removes area from a positive integrand, so the derivative must be negative.

Problem 4

Let $G(x) = \displaystyle\int_x^{2x} e^{-t}\,dt$ for $x \ge 0$ Find $G'(x)$ and the $x$ at which $G$ is largest.

$G'(x) = e^{-2x}\cdot2 - e^{-x}\cdot1$ Upper limit $2x$ with derivative $2$ minus the lower limit $x$ with derivative $1$
$G'(x) = 0 \iff 2e^{-2x} = e^{-x} \iff e^{-x} = \tfrac12$ Divide by $e^{-x} > 0$
$x = \ln2$ $G' > 0$ for $x < \ln2$ and $G' < 0$ after, so this is the maximum.
$G(\ln2) = e^{-\ln2} - e^{-2\ln2} = \tfrac12 - \tfrac14$ $G(x) = e^{-x} - e^{-2x}$ by Part 2.
$G'(x) = 2e^{-2x} - e^{-x}$ $G$ is largest at $x = \ln2 \approx 0.693$ where $G = \tfrac14$ $G(0) = 0$ and $G \to 0$ as $x \to \infty$ the interval $[x, 2x]$ starts empty and ends where $e^{-t}$ is tiny.

Problem 5

Let $F(x) = \displaystyle\int_0^x (t^2 - 4t + 3)\,dt$ Find the local maximum and minimum values of $F$

$F'(x) = x^2 - 4x + 3 = (x - 1)(x - 3)$ Part 1. The critical points of $F$ are the zeros of its integrand.
$F''(x) = 2x - 4$ so $F''(1) = -2$ and $F''(3) = 2$ Local maximum at $1$ local minimum at $3$
$F(x) = \tfrac{x^3}3 - 2x^2 + 3x$ Part 2; the lower limit $0$ adds nothing.
$F(1) = \tfrac13 - 2 + 3 = \tfrac43$ and $F(3) = 9 - 18 + 9 = 0$ The values of $F$ not of the integrand.
Local maximum $\tfrac43$ at $x = 1$ local minimum $0$ at $x = 3$ $F$ climbs while the integrand is positive, on $(0, 1)$ and loses exactly that area again on $(1, 3)$ where the integrand is negative.

Problem 6

Evaluate $\displaystyle\int_1^4 \Big(2x + \frac{1}{\sqrt x}\Big)\,dx$

An antiderivative is $x^2 + 2\sqrt x$ $\int x^{-1/2}\,dx = \tfrac{x^{1/2}}{1/2} = 2\sqrt x$
$\big(16 + 2\cdot2\big) - \big(1 + 2\cdot1\big) = 20 - 3$ $F(4) - F(1)$ The integrand is continuous on $[1, 4]$
$17$ Term by term: $\int_1^4 2x\,dx = 15$ and $\int_1^4 x^{-1/2}\,dx = 2$

Problem 7

Evaluate $\displaystyle\int_0^3 \lvert x - 1\rvert\,dx$

$\lvert x - 1\rvert = 1 - x$ on $[0, 1]$ and $x - 1$ on $[1, 3]$ Split where the inside changes sign.
$\displaystyle\int_0^1 (1 - x)\,dx = \Big[x - \tfrac{x^2}2\Big]_0^1 = \tfrac12$ The first piece.
$\displaystyle\int_1^3 (x - 1)\,dx = \Big[\tfrac{x^2}2 - x\Big]_1^3 = \tfrac32 - \big(-\tfrac12\big) = 2$ The second piece.
$\tfrac52$ Two triangles: one with legs $1$ and $1$ one with legs $2$ and $2$

Problem 8

$\Phi$ is the standard normal CDF. Find $\dfrac{d}{dx}\Phi(2x - 1)$ and its value at $x = \tfrac12$

$\Phi(2x - 1) = \displaystyle\int_{-\infty}^{2x - 1}\varphi(t)\,dt$ A CDF is an integral with a variable upper limit.
$\dfrac{d}{dx}\Phi(2x - 1) = \varphi(2x - 1)\cdot2$ Part 1 with the chain rule; the lower limit $-\infty$ does not move.
At $x = \tfrac12$ $2\varphi(0) = \dfrac{2}{\sqrt{2\pi}}$ $\varphi(0) = \tfrac1{\sqrt{2\pi}}$
$2\varphi(2x - 1)$ at $x = \tfrac12$ it is $\sqrt{2/\pi} \approx 0.798$ This is the density of $X = \tfrac{Z + 1}2$ $P(X \le x) = \Phi(2x - 1)$ and differentiating a CDF gives the density. It is the normal density with mean $\tfrac12$ and standard deviation $\tfrac12$

Problem 9

The waiting time $T$ has density $p(t) = 2e^{-2t}$ for $t \ge 0$ Find its CDF $F(x) = P(T \le x)$ check that $F' = p$ and find the median.

$F(x) = \displaystyle\int_0^x 2e^{-2t}\,dt = \Big[-e^{-2t}\Big]_0^x$ The density is $0$ for $t < 0$ so the integral starts at $0$
$F(x) = 1 - e^{-2x}$ for $x \ge 0$ $-e^{-2x} - (-1)$ $F(0) = 0$ and $F \to 1$
$F'(x) = 2e^{-2x} = p(x)$ Part 1, confirmed by differentiating the closed form.
$1 - e^{-2m} = \tfrac12 \iff e^{-2m} = \tfrac12 \iff m = \tfrac{\ln2}2$ The median is where the CDF reaches $\tfrac12$
$F(x) = 1 - e^{-2x}$ median $\tfrac{\ln2}2 \approx 0.347$ The mean is $\tfrac12$ larger than the median: the long right tail pulls it up.

Problem 10

Find $\displaystyle\lim_{x\to0}\frac{1}{x^3}\int_0^x \sin(t^2)\,dt$

As $x \to 0$ both $\int_0^x \sin(t^2)\,dt$ and $x^3$ go to $0$ The form is $\tfrac00$ so L'Hôpital's rule applies.
$\displaystyle\lim_{x\to0}\frac{\sin(x^2)}{3x^2}$ Differentiate the top by Part 1 and the bottom by the power rule.
$\dfrac{\sin(x^2)}{x^2} \to 1$ With $u = x^2 \to 0$ $\tfrac{\sin u}u \to 1$
$\tfrac13$ Near $0$ $\sin(t^2) \approx t^2$ and $\int_0^x t^2\,dt = \tfrac{x^3}3$ the same answer without the rule.

Where this goes wrong

1. Leaving out the chain-rule factor

Part 1 is learned as "put the upper limit in", which is the whole rule only when the limit is $x$

$\dfrac{d}{dx}\displaystyle\int_1^{x^3}\ln t\,dt$ $x > 0$ Right so far: Problem 2.
“Replace $t$ with the upper limit.”The habit that causes the mistake: the limit is $x^3$ not $x$ so it moves $3x^2$ times as fast.
$\ln(x^3) = 3\ln x$ so at $x = 2$ the derivative is $3\ln2 \approx 2.079$ The edge of the area moves at the rate $(x^3)' = 3x^2$ and the derivative is $9x^2\ln x$ which is $36\ln2 \approx 24.95$ at $x = 2$ twelve times as much. Differentiating the closed form $3x^3\ln x - x^3 + 1$ shows the factor.

2. Dropping the sign of a variable lower limit

In most problems $x$ is the upper limit, so the minus sign is easy to forget.

$\dfrac{d}{dx}\displaystyle\int_x^5 \sqrt{1 + t^4}\,dt$ Right so far: Problem 3.
“Put $x$ into the integrand.”The habit that causes the mistake: Part 1 applied without first moving $x$ to the upper limit.
The derivative is $\sqrt{1 + x^4}$ so $\sqrt2 \approx 1.414$ at $x = 1$ $\int_x^5 = -\int_5^x$ so the derivative is $-\sqrt{1 + x^4}$ which is $-1.414$ at $x = 1$ As $x$ increases, the interval $[x, 5]$ gets shorter and the integrand is positive, so the integral is decreasing.

3. Subtracting the integrand at the lower limit

Part 2 is "top minus bottom", and the pattern carries over to Part 1, where it does not belong.

$F(x) = \displaystyle\int_1^x (t^3 - 2t)\,dt$ Right so far: Problem 1.
“The derivative is $f$ at the top minus $f$ at the bottom.”The habit that causes the mistake: Part 2's subtraction copied into Part 1.
$F'(x) = (x^3 - 2x) - (1 - 2) = x^3 - 2x + 1$ The lower limit is a constant, and it contributes a constant, $\tfrac34$ to $F$ which differentiates to $0$ $F'(x) = x^3 - 2x$ at $x = 2$ that is $4$ not $5$ Only a lower limit that moves with $x$ contributes to the derivative.

4. Integrating an absolute value without splitting

The bars look like decoration once the integral is set up.

$\displaystyle\int_0^3 \lvert x - 1\rvert\,dx$ Right so far: Problem 7.
“For most of the interval $x - 1 \ge 0$ so use $x - 1$ ”The habit that causes the mistake: one formula used on both sides of $x = 1$
$\displaystyle\int_0^3 (x - 1)\,dx = \tfrac92 - 3 = \tfrac32$ On $[0, 1]$ $x - 1$ is negative, so the triangle there, of area $\tfrac12$ is subtracted instead of added. Split at $x = 1$ $\tfrac12 + 2 = \tfrac52$ The integral of a function that is never negative is a total area, and here $\tfrac32 = 2 - \tfrac12$ is a net one.

5. Reporting the integrand's value instead of F's

$F$ and its integrand share the same critical points, which makes it easy to evaluate the wrong one.

$F(x) = \displaystyle\int_0^x (t^2 - 4t + 3)\,dt$ local maximum at $x = 1$ Right so far: Problem 5, step 2.
“Plug $x = 1$ into the function.”The habit that causes the mistake: the function on the page is the integrand.
The local maximum value is $1 - 4 + 3 = 0$ $0$ is the integrand at $x = 1$ it is $F'(1)$ which is $0$ at every critical point. The value of $F$ is the area accumulated up to $1$ $F(1) = \tfrac13 - 2 + 3 = \tfrac43$ The local minimum value is $F(3) = 0$ which happens to be $0$ for a different reason.

Print this set: fundamental-theorem-of-calculus.pdf (problems, answers, and worked solutions on separate pages).

Read the guide: The fundamental theorem of calculus: derivatives of integrals, and the chain rule on the limit