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AP Calc AB · Mathematics
Integrals · Definite integral

The Fundamental Theorem of Calculus

Before you start, make sure you know: Continuity of functions (Precalculus)
Before you start, make sure you know: Continuity of a function (Precalculus)

Emma's family is driving from Houston to Dallas. Her phone logs the car's speed every second, and the odometer logs the total miles. If you only had the speed log, could you recover how far they drove? If you only had the odometer, could you recover the speed? Speed is the derivative of position, and distance is the area under the speed graph. The Fundamental Theorem of Calculus (FTC) says these two ideas undo each other. It also gives you a fast way to compute areas.

Accumulation functions and FTC Part 1

Accumulation function

Fix the lower limit $a$ and let the upper limit move. For each $x$, the integral $\int_a^x f(t)\,dt$ is the signed area collected from $a$ up to $x$. Every value of $x$ gives one number, so this is a new function of $x$.

Animation of an accumulation function: the dashed vertical line slides right and then left along the axis, and the shaded area under the curve from a to x grows and shrinks with it. The label F(x) = area up to x moves along with the edge. a x F(x) = area up to x

Let's compute one by hand. For $F(x)=\int_0^x 3t\,dt$, the region under $y=3t$ from $0$ to $x$ is a triangle with base $x$ and height $3x$. So $F(x)=\tfrac12 \cdot x \cdot 3x=\tfrac{3x^2}{2}$ and $F(4)=24$. Notice that $F'(x)=3x$, which is exactly the original function. That is not a coincidence.

FTC Part 1

On the AP exam, state the hypothesis: "since $f$ is continuous, by the FTC, $F'(x)=f(x)$."

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Find the value of the derivative of the following integral function at the point $x = 2$: $$F(x) = \int_{0}^{x} \sqrt{t^3 + 1} \, dt$$ Explain which mathematical theorem…

Find the value of the derivative of the following integral function at the point $x = 2$:

$$F(x) = \int_{0}^{x} \sqrt{t^3 + 1} \, dt$$

Explain which mathematical theorem you are using to solve the problem without needing to calculate the antiderivative.

Solve
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A video game programmer is designing the trajectory of a projectile. The speed of the object as a function of time $t$ is given by the expression…

A video game programmer is designing the trajectory of a projectile. The speed of the object as a function of time $t$ is given by the expression $v(t) = \int_1^t (3x^2 - 4x) \, dx$.

Apply the Fundamental Theorem of Calculus to find the expression for the instantaneous acceleration $a(t)$, knowing that acceleration is the derivative of speed with respect to time ($a(t) = v'(t)$).

Solve

When the limits are functions of x

What if the upper limit is $x^2$ instead of $x$? Then $F$ is a composition: the accumulation function evaluated at $g(x)$. The chain rule gives an extra factor.

Pattern for differentiating an integral with variable limits: plug the upper limit into f and multiply by its derivative, then subtract the same for the lower limit. d/dx [ area from g(x) up to h(x) ] = f(h(x)) · h′(x) − f(g(x)) · g′(x) top limit in, times its derivative; minus the bottom one
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Calculate the derivative function $F'(x)$ of the following integral function using the Fundamental Theorem of Calculus and the chain rule: $$F(x) = \int_{1}^{x^2}…

Calculate the derivative function $F'(x)$ of the following integral function using the Fundamental Theorem of Calculus and the chain rule:

$$F(x) = \int_{1}^{x^2} \frac{\ln(t)}{t + 1} \, dt$$

State the domain of definition of the function $F(x)$ so that the integral is well-defined.

Solve
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Let the function $f(t) = \frac{\ln(t)}{t^2}$ be defined for $t > 0$. A new integral function $G(x)$ is defined as: $$G(x) = \int_{x}^{x^2} f(t) \, dt$$ a) Applying the…

Let the function $f(t) = \frac{\ln(t)}{t^2}$ be defined for $t > 0$. A new integral function $G(x)$ is defined as:

$$G(x) = \int_{x}^{x^2} f(t) \, dt$$
  1. a) Applying the Fundamental Theorem of Calculus and the chain rule, obtain the expression for the derivative $G'(x)$.
  2. b) Determine the intervals of increase and decrease of $G(x)$ in its domain.
  3. c) Find the value of $x > 1$ for which the function $G(x)$ reaches a relative maximum.
t y 1 x x² f(t) G(x)
Solve

Increasing, decreasing, and extrema of F

Since $F'(x)=f(x)$, the graph of $f$ tells you everything about the shape of $F$, even when you cannot compute $F$.

What you see on $f$ What it means for $F$
$f>0$ $F$ is increasing
$f<0$ $F$ is decreasing
$f$ changes from $-$ to $+$ at $c$ $F$ has a relative minimum at $c$
$f$ changes from $+$ to $-$ at $c$ $F$ has a relative maximum at $c$
$f$ increasing (so $F''=f'>0$) $F$ is concave up
$f'$ changes sign at $c$ $F$ has a point of inflection at $c$
Top: a graph of f that is negative then positive, crossing the axis at x = 1. Bottom: the matching accumulation function F, which decreases to a minimum at x = 1 and then increases. f = F′ F 1 f < 0 on the left: F decreasing f > 0 on the right: F increasing, minimum at x = 1

The sign of $F$ itself depends on the integrand and on the direction you travel from $a$. Moving right through a negative integrand makes $F$ negative. Moving left through a negative integrand makes $F$ positive, because $\int_x^a f = -\int_a^x f$.

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Given the integral function $H(x) = \int_{0}^{x} \frac{t-1}{t^2+4} \, dt$: a) Find the intervals of increase and decrease of $H(x)$. b) Determine the coordinates of the…

Given the integral function $H(x) = \int_{0}^{x} \frac{t-1}{t^2+4} \, dt$:

  1. a) Find the intervals of increase and decrease of $H(x)$.
  2. b) Determine the coordinates of the relative extrema (maxima or minima) of $H(x)$ and classify them.
Solve
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Let the function be defined by the integral $F(x) = \int_{0}^{x} \frac{t - 1}{e^t + 1} \, dt$. a) Find the critical points of $F(x)$ and classify them using the first…

Let the function be defined by the integral $F(x) = \int_{0}^{x} \frac{t - 1}{e^t + 1} \, dt$.

  1. a) Find the critical points of $F(x)$ and classify them using the first derivative test.
  2. b) Study the curvature of $F(x)$ and determine if it has any points of inflection.
  3. c) Is there any value of $x$ for which $F(x)$ is negative? Reason your answer based on the sign of the integrand and the limits of integration.
Solve

Tangent lines, concavity, and motion

An accumulation function comes with a free anchor: $F(a)=0$. To write the tangent line at $x=c$, you need the point $(c, F(c))$ and the slope $F'(c)=f(c)$. Often $c=a$, which makes $F(c)=0$.

For the second derivative, differentiate the integrand: $F''(x)=f'(x)$.

Motion is the same idea. If $v(t)=\int_0^t g(z)\,dz$ is a velocity, then the acceleration is $v'(t)=g(t)$. To find where the acceleration is largest on $[0,4]$, study $a(t)=g(t)$ itself: compute $a'(t)$ and check its sign and the endpoints.

The graph above is a decreasing function. A decreasing function has a negative derivative, so its maximum on $[0,4]$ is at the left endpoint.

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Let the function $G(x)$ be defined by: $$G(x) = \int_{0}^{x} (t^2 - 4)e^{t} \, dt$$ Find the equation of the tangent line to the graph of $G(x)$ at the point with…

Let the function $G(x)$ be defined by:

$$G(x) = \int_{0}^{x} (t^2 - 4)e^{t} \, dt$$

Find the equation of the tangent line to the graph of $G(x)$ at the point with x-coordinate $x = 0$.

Note: Remember that to find the y-coordinate of the point of tangency you must calculate $G(0)$ and for the slope you must apply the Fundamental Theorem of Calculus.

Solve
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A team of engineers is designing the trajectory of a racing drone. The vertical speed of the drone in the time interval $[0, 4]$ seconds is given by the function…

A team of engineers is designing the trajectory of a racing drone. The vertical speed of the drone in the time interval $[0, 4]$ seconds is given by the function $v(t) = \int_{0}^{t} \frac{1}{1 + z^4} \, dz$.

  1. a) Without calculating the integral explicitly, justify why the drone is always rising or maintaining the same height.
  2. b) Calculate the vertical acceleration of the drone at the instant $t = 1$ second.
  3. c) At what instant is the vertical acceleration maximum? Justify your answer by studying the second derivative of the position.
Solve
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Consider the integral function $L(x) = \int_{1}^{x} \frac{\sqrt{1 + t^3}}{t} \, dt$ defined for $x \geq 1$. a) Show that $L(x)$ is a strictly increasing function in its…

Consider the integral function $L(x) = \int_{1}^{x} \frac{\sqrt{1 + t^3}}{t} \, dt$ defined for $x \geq 1$.

  1. a) Show that $L(x)$ is a strictly increasing function in its domain.
  2. b) It is known that the graph of $L(x)$ has a tangent line at the point with x-coordinate $x = 2$ whose equation is $y = mx + n$. Calculate the value of the slope $m$.
  3. c) Determine the acceleration of the slope's growth (second derivative) at that same point.
x y 0 1 2 L(2) P(2, L(2)) L(x) y = mx + n Δx = 1 Δy = m
Solve

FTC Part 2: evaluating definite integrals

Part 1 says an accumulation function is an antiderivative. This is the key to computing definite integrals without adding up rectangles.

Some problems call this Barrow's Rule. It is the same as FTC Part 2, the "evaluation" part of the theorem, sometimes called the Net Change Theorem when $f$ is a rate.

Scheme of FTC Part 2: the definite integral from a to b equals G(b) minus G(a), where G is an antiderivative of f. ∫ₐᵇ f(x)dx = G(b) − G(a) 1. find an antiderivative G of f 2. evaluate at b and at a, then subtract

When $f$ is a rate of change, Part 2 says: $\displaystyle\int_a^b f'(x)\,dx=f(b)-f(a)$. The integral of a rate is the net change. That is how the odometer is recovered from the speed log.

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The electricity consumption of a house during the first hour of the morning is modeled by the function $f(x) = e^x + x$, where $x$ is the time in hours. Calculate the…

The electricity consumption of a house during the first hour of the morning is modeled by the function $f(x) = e^x + x$, where $x$ is the time in hours. Calculate the total accumulated consumption, represented by the definite integral:

$$\int_{0}^{1} (e^x + x) \, dx$$
Solve
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A solar thermal panel has a design whose upper cross-section is bounded by the function $f(x) = 3x^2 + 2$, where $x$ is measured in feet. Calculate the value of the…

A solar thermal panel has a design whose upper cross-section is bounded by the function $f(x) = 3x^2 + 2$, where $x$ is measured in feet. Calculate the value of the definite integral in the interval $[0, 2]$ using Barrow's Rule.

0 1 2 x 2 14 y f(x) = 3x² + 2 ∫₀² f(x) dx
Solve
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Determine the value of the positive real parameter $a$ knowing that the area bounded by the curve $f(x) = 3x^2$, the $OX$-axis and the vertical lines $x = 0$ and $x = a$…

Determine the value of the positive real parameter $a$ knowing that the area bounded by the curve $f(x) = 3x^2$, the $OX$-axis and the vertical lines $x = 0$ and $x = a$ is exactly equal to 27 square units. Use Barrow's Rule for this.

X Y 0 a f(a) f(x) = 3x² Area = 27 u²
Solve

Properties, symmetry, and substitution

Properties you will use all the time

Property Statement
Constant multiple $\int_a^b k f(x)\,dx = k\int_a^b f(x)\,dx$
Sum $\int_a^b [f(x)+g(x)]\,dx = \int_a^b f+\int_a^b g$
Additivity $\int_a^c f = \int_a^b f + \int_b^c f$
Reversing limits $\int_b^a f = -\int_a^b f$
Same limits $\int_a^a f = 0$

Symmetry

For an interval $[-a,a]$: if $f$ is odd ($f(-x)=-f(x)$), the signed area on the left cancels the signed area on the right, so $\int_{-a}^a f=0$. If $f$ is even, $\int_{-a}^a f=2\int_0^a f$.

Graph of an odd function on the interval from -2 to 2. The region above the axis on the right and the region below the axis on the left have equal size and opposite signs, so they cancel. area + area − −2 2

Substitution in a definite integral

When you substitute $u=g(x)$, change the limits too, so you never have to go back to $x$.

Practice★☆☆Not started
Given a continuous function $f(x)$ in the interval $[1, 8]$, the values of the following definite integrals are known: $\int_1^5 f(x) \, dx = 7$…

Given a continuous function $f(x)$ in the interval $[1, 8]$, the values of the following definite integrals are known:

  • $\int_1^5 f(x) \, dx = 7$

  • $\int_5^8 f(x) \, dx = -3$

Use the properties of the definite integral to calculate the value of $\int_1^8 2 \cdot f(x) \, dx$.

Solve
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Without performing complex integration calculations, justify with reasoning why the value of the following integral is equal to $0$, based on the properties of odd…

Without performing complex integration calculations, justify with reasoning why the value of the following integral is equal to $0$, based on the properties of odd functions and the symmetry of the interval:

$$\int_{-2}^{2} (x^5 + 3x^3 - x) \, dx$$ x y −2 2 0 -A +A f(x) = x⁵ + 3x³ - x f(-x) = -f(x) ∫₋₂² f(x)dx = 0 -A + A = 0
Solve
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Let $f(x)$ be a continuous function in $\mathbb{R}$ such that $\int_{0}^{1} f(x) \, dx = 2$. Find the value of the constant $k$ so that the following equality holds,…

Let $f(x)$ be a continuous function in $\mathbb{R}$ such that $\int_{0}^{1} f(x) \, dx = 2$. Find the value of the constant $k$ so that the following equality holds, knowing that the area under the curve on the left has been shifted and scaled:

$$\int_{0}^{2} f\left(\frac{x}{2}\right) \, dx + \int_{1}^{2} x \cdot f(x^2 - 1) \, dx = k$$

Use the integration by substitution method in the definite integrals to relate them to the initial data.

Solve

Signed value, area, and displacement

$\int_a^b f(x)\,dx$ is a signed area: regions below the $x$-axis count as negative. The geometric area is $\int_a^b |f(x)|\,dx$. The same difference shows up in motion:

  • Displacement $=\int_a^b v(t)\,dt$ (the net change in position).

  • Total distance $=\int_a^b |v(t)|\,dt$.

Graph of y = x squared minus 1 from x = 0 to x = 1.5. The region between the curve and the x-axis from 0 to 1 is below the axis, so its signed value is negative 2/3 while its area is 2/3. y = x² − 1 signed: −2/3 area: 2/3 1
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Calculate the area of the plane region bounded by the graph of the function $f(x) = \sin(x)$, the x-axis $OX$ and the vertical lines $x = 0$ and $x = \pi$. Sketch the…

Calculate the area of the plane region bounded by the graph of the function $f(x) = \sin(x)$, the x-axis $OX$ and the vertical lines $x = 0$ and $x = \pi$. Sketch the region before applying Barrow's Rule.

X Y 0 1 π/2 π f(x) = sin(x) x=0 x=π S
Solve
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Determine the area of the region enclosed between the curve $f(x) = x^2 - 4$ and the $OX$-axis in the interval $[0, 2]$. Remember that if the function is negative in the…

Determine the area of the region enclosed between the curve $f(x) = x^2 - 4$ and the $OX$-axis in the interval $[0, 2]$. Remember that if the function is negative in the interval, the area corresponds to the absolute value of the definite integral.

x y −2 −1 1 2 3 1 −1 −2 −3 −4 0 f(x) = x² - 4 (0, −4) Area
Solve
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The speed of an electric vehicle prototype during the first 4 seconds of a test is given by the function $v(t) = t^2 - 3t + 4$ (in ft/s). a) Calculate the total distance…

The speed of an electric vehicle prototype during the first 4 seconds of a test is given by the function $v(t) = t^2 - 3t + 4$ (in ft/s).

  1. a) Calculate the total distance traveled by the vehicle between the instants $t = 1$ s and $t = 3$ s applying Barrow's Rule.
  2. b) Justify whether the vehicle stops at any point during that interval.
Solve

Splitting the interval: absolute value and piecewise functions

FTC Part 2 uses one antiderivative on one formula. If the formula changes inside the interval, split the integral where it changes.

  1. Absolute value: find the zeros of the expression inside $|\ \cdot\ |$ on the interval, and check the sign on each piece. Where it is negative, integrate $-f$.

  2. Piecewise function: first check continuity at the break point, then integrate each formula on its own piece.

Graph of y = absolute value of x squared minus 3x from 0 to 4. The region from 0 to 3 has area 9/2 and the region from 3 to 4 has area 11/6. A dashed line marks the split at x = 3. 9/2 11/6 3 y = |x² − 3x|

Now a subtle point about accumulation functions with an absolute value, such as $H(x)=\int_0^x |t^2-4|\,dt$. The integrand $|t^2-4|$ has corners at $t=\pm2$, but it is still continuous. FTC Part 1 only requires a continuous integrand, so $H'(x)=|x^2-4|$ exists at every point, including $x=2$ (where it equals $0$). The corner of the integrand does not create a corner in $H$. As a piecewise function, $H$ must be defined using the antiderivative on each piece, with the pieces joined at the break point.

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Given the function $f(x) = |x^2 - 2x|$, calculate the value of the definite integral in the interval $[0, 3]$ using Barrow's Rule: $$\int_0^3 |x^2 - 2x| \, dx$$ Briefly…

Given the function $f(x) = |x^2 - 2x|$, calculate the value of the definite integral in the interval $[0, 3]$ using Barrow's Rule:

$$\int_0^3 |x^2 - 2x| \, dx$$

Briefly explain why it is necessary to split the interval of integration to solve this exercise.

0 1 2 3 −1 1 2 3 x y f(x) = |x² - 2x| y = x² - 2x A₁ A₂
Solve
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Calculate the value of the definite integral of the following piecewise function in the interval $[0, 2]$: $$f(x) = \begin{cases} e^{2x} & \text{if } 0 \le x \le 1 \\…

Calculate the value of the definite integral of the following piecewise function in the interval $[0, 2]$:

$$f(x) = \begin{cases} e^{2x} & \text{if } 0 \le x \le 1 \\ \frac{e^2}{x} & \text{if } 1 < x \le 2 \end{cases}$$

Check beforehand if the function is continuous at $x = 1$ to ensure that it is integrable.

0 1 2 x 1 e² y f(x) = e²ˣ f(x) = e²/x (1, e²) A₁ A₂
Solve
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Given the function $f(x) = x^2 - 4$, consider the integral function $H(x) = \int_{0}^{x} |f(t)| \, dt$. a) Express $H(x)$ as a piecewise function for values of…

Given the function $f(x) = x^2 - 4$, consider the integral function $H(x) = \int_{0}^{x} |f(t)| \, dt$.

  1. a) Express $H(x)$ as a piecewise function for values of $x \in [0, 5]$.
  2. b) Study the differentiability of $H(x)$ at the point $x = 2$. Is the Fundamental Theorem of Calculus directly applicable at that point?
  3. c) Use Barrow's Rule to calculate the exact value of $H(3)$.
t y A₁ A₂ 2 3 4 −4 y = |t² - 4| y = t² - 4
Solve

Area between two curves

If $f(x)\ge g(x)$ on $[a,b]$, the area between the graphs is $\int_a^b [f(x)-g(x)]\,dx$. Find where the graphs meet first (set $f=g$), because those points are the limits. To decide which graph is on top, test a point between them.

The line y = 2x and the parabola y = x squared meet at x = 0 and x = 2. The shaded region between them is the area, with the line on top. y = 2x y = x² 0 2
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The logo of a new streetwear brand is bounded by the graphs of the functions $f(x) = \operatorname{sen}(\pi x)$ and $g(x) = ax$, with $a > 0$, in the first quadrant. a)…

The logo of a new streetwear brand is bounded by the graphs of the functions $f(x) = \operatorname{sen}(\pi x)$ and $g(x) = ax$, with $a > 0$, in the first quadrant.

  1. a) Determine the value of $a$ so that the line $y = ax$ passes through the relative maximum of the function $f(x)$ in the interval $[0, 1]$.
  2. b) For the value of $a$ found, calculate the area of the region enclosed between both graphs using Barrow's Rule.
0 0.5 1 1 x y f(x) = sin(πx) g(x) = ax Maximum Area
Solve

Average value and the Mean Value Theorem for Integrals

If you drive $140$ miles in $2$ hours, your average speed is $70$ mph. The same idea works for any continuous function.

Geometrically, the rectangle with width $b-a$ and height $f_{\text{avg}}$ has the same area as the region under the curve. The value $c$ is the $x$-coordinate where the curve crosses the top of that rectangle.

Graph of y = x squared on the interval 0 to 3 with a horizontal dashed line at the average value 3. The curve crosses that line at x equal to the square root of 3. avg = 3 c = √3 0 3
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Find the mean value of the function $f(x) = \sqrt{x+1}$ in the interval $[0, 3]$ and determine the point $c \in (0, 3)$ whose existence is guaranteed by the Mean Value…

Find the mean value of the function $f(x) = \sqrt{x+1}$ in the interval $[0, 3]$ and determine the point $c \in (0, 3)$ whose existence is guaranteed by the Mean Value Theorem for Definite Integrals. Interpret the result geometrically.

0 1 2 3 c 1 2 f(c) f(x) = √(x+1)
Solve

Limits of integrals with L'Hôpital's Rule

An accumulation function can show up in a limit. If the integral has a variable upper limit and starts at the same value as the limit point, the numerator tends to $0$. Together with a denominator that tends to $0$, you have a $\frac00$ form, and L'Hôpital's Rule applies. The FTC differentiates the numerator for you.

Three-step plan for a limit with an integral: check that it is a 0 over 0 form, differentiate the top with the FTC and the bottom normally, then take the limit and repeat if needed. 1. Check 0/0(integral at 0 is 0) 2. L'Hôpital:top by FTC 3. Take the limit(repeat if 0/0)
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Calculate the value of the following limit using L'Hôpital's Rule and the Fundamental Theorem of Calculus: $$\lim_{x \to 0} \frac{\int_{0}^{x^2} \left( e^{t^2} - 1…

Calculate the value of the following limit using L'Hôpital's Rule and the Fundamental Theorem of Calculus:

$$\lim_{x \to 0} \frac{\int_{0}^{x^2} \left( e^{t^2} - 1 \right) \, dt}{x^6}$$

Explain each step with reasoning, justifying the differentiation of the integral function.

Solve

Back to the road trip: the speed log $v(t)$ gives the distance by $\int_a^b v(t)\,dt=s(b)-s(a)$ (Part 2), and the odometer gives the speed back because $s'(t)=v(t)$ (Part 1). Distance and speed are two views of the same relationship.

Summary

See all the exercises in this section