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Secondary 6 · Mathematics
Integrals · Definite integral

Fundamental Theorem of Integral Calculus and Barrow's Rule

The electric bill for a house in Santiago depends on accumulated energy: if you know the power the house uses at each instant, the total energy is the area under the graph of that power. Calculating that area by adding up rectangles is slow and only gives approximations. The fundamental theorem of calculus connects area (integral) with slope (derivative) and gives you an exact shortcut: Barrow's rule. At the end we come back to this situation.

The definite integral as area

To measure the area under a curve $y=f(x)$ between $x=a$ and $x=b$, with $f\ge0$, we divide $[a,b]$ into $n$ pieces of equal width $\Delta x=\dfrac{b-a}{n}$ and build a rectangle on each piece. The sum of the areas of the rectangles approximates the area we want.

Four rectangles under the parabola y = x² between x = 0 and x = 1, with the height of each rectangle taken at its right endpoint; the rectangles approximate the area. 0 1 y = x²

With $n=4$ and heights taken at the right endpoint of each piece, for $f(x)=x^2$ on $[0,1]$:

$$S_4=\frac14\left(\frac1{16}+\frac4{16}+\frac9{16}+\frac{16}{16}\right)=\frac{30}{64}\approx 0.47$$

With more rectangles the approximation gets better. The value the sum approaches as $n\to\infty$ is the definite integral.

The sign. Each rectangle has height $f(x_i)$. If $f$ is negative, that height is negative and the "area" it contributes counts with a minus sign. That is why the integral is a signed area: what lies above the $x$-axis adds and what lies below subtracts.

Splitting where the sign changes. The geometric area (always positive) is obtained by splitting the region at the zeros of $f$ and adding the absolute values of each piece.

Properties of the definite integral

These rules come from thinking of the integral as signed area, and they let you combine integrals you already know.

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Given a function $f(x)$ continuous on the interval $[1, 8]$, the values of the following definite integrals are known: $\int_1^5 f(x) \, dx = 7$…

Given a function $f(x)$ continuous on 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 $\int_1^8 2 \times f(x) \, dx$.

Solve

The function defined by an integral

Fix the lower endpoint $a$ and let the upper one vary. For each $x$, the number $F(x)=\int_a^x f(t)\,dt$ is the area accumulated from $a$ to $x$. So $F$ is a new function: it assigns an area to each $x$.

Graph of a curve above the x-axis with the area shaded from a to x, representing F(x), the accumulated area. a x F(x) = area y = f(t)

The fundamental theorem of calculus

How does the accumulated area change when $x$ moves forward a little? Here is the key piece of integral calculus.

Diagram: integrating takes you from f to F, differentiating takes you from F to f; they are inverse operations. F(x) = ∫ f(t) dt F'(x) = f(x) differentiate → ← integrate integrating and differentiating undo each other

When the upper limit is not $x$ but a function of $x$, the chain rule also comes into play. And if both limits depend on $x$, the integral is split into two integrals.

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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 velocity 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 velocity 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 velocity with respect to time ($a(t) = v'(t)$).

Solve
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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

Studying an integral function without calculating the integral

Sometimes the integral has no simple antiderivative. It doesn't matter: $F'(x)=f(x)$ already tells you almost everything about $F$.

  • Where $f>0$, $F$ increases. Where $f<0$, $F$ decreases.

  • The critical points of $F$ are the zeros of $f$. If $f$ goes from $-$ to $+$, there is a minimum; from $+$ to $-$, a maximum.

  • The curvature comes from $F''(x)=f'(x)$: if $f'$ changes sign, $F$ has an inflection point.

  • The tangent line at $x=x_0$ has slope $F'(x_0)=f(x_0)$ and passes through $(x_0,F(x_0))$.

Two aligned graphs with labeled axes: on top, f(t) crosses the t-axis from negative to positive at t = 3; below, F decreases down to that point, has a minimum, and then increases. t x f(t): from − to + F(x): minimum at x = 3 3 3
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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 gradient you must apply the Fundamental Theorem of Calculus.

Solve
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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 local extrema (maximums or minimums) of $H(x)$ and classify them.
Solve
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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
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A team of engineers is designing the trajectory of a racing drone. The vertical velocity 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 velocity 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 explicitly calculating the integral, justify why the drone is always rising or remaining at the same height.
  2. b) Calculate the vertical acceleration of the drone at time $t = 1$ 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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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 local maximum.
t y 1 x x² f(t) G(x)
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 concavity 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
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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 gradient $m$.
  3. c) Determine the rate of change of the gradient (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

Barrow's rule

Now the promised shortcut. If you know an antiderivative of $f$, the definite integral comes down to evaluating and subtracting.

Diagram of Barrow's rule: the integral from a to b is the antiderivative evaluated at b minus the one evaluated at a. ∫ₐᵇ f(x) dx = G(b) − G(a) 1. find an antiderivative G of f 2. evaluate at b and at a, and subtract

Signed areas

If $f$ is negative on the interval, the integral comes out negative. The geometric area is always positive: it is the absolute value of that integral. If $f$ changes sign, you must split at the zeros and add the absolute values.

Area between two curves

If two curves enclose a region, its area is the integral of the upper curve minus the lower one. The points where they intersect are the limits of integration.

$$A=\int_a^b \big(f(x)-g(x)\big)\,dx\qquad\text{with } f(x)\ge g(x)\text{ on }[a,b]$$

Why? At each $x$, the height of the region is the vertical distance $f(x)-g(x)$. Adding up those heights is integrating the difference.

Region enclosed between the line y = x and the parabola y = x² between x = 0 and x = 1; the line is on top. 0 1 y = x y = x²
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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 meters. 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 meters. Calculate the value of the definite integral on 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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The electricity consumption of a household during the first hour of the morning is modelled by the function $f(x) = e^x + x$, where $x$ is the time in hours. Calculate…

The electricity consumption of a household during the first hour of the morning is modelled 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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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

You can do it on paper and send a photo with your answer.

Solve
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Determine the area of the region enclosed between the curve $f(x) = x^2 - 4$ and the $OX$ axis on the interval $[0, 2]$. Remember that if the function is negative on the…

Determine the area of the region enclosed between the curve $f(x) = x^2 - 4$ and the $OX$ axis on the interval $[0, 2]$. Remember that if the function is negative on the interval, the area corresponds to the modulus 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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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 units of area. Use Barrow's Rule to do this.

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

Accumulated quantities: distance traveled

If $v(t)$ is the velocity of a car, then $x'(t)=v(t)$, and by Barrow the change in position between $t_1$ and $t_2$ is an integral. The same idea works for any rate: energy consumption, cost per unit produced, water flowing into a tank.

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The velocity of a prototype electric vehicle during the first 4 seconds of a test is given by the function $v(t) = t^2 - 3t + 4$ (in m/s). a) Calculate the total…

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

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

Absolute value and piecewise functions

When the formula for $f$ changes inside the interval, you split the integral at the points of change and apply Barrow on each piece. First, check that the function is continuous at the joining point.

Piecewise graph: the line y = 2x from x = 0 to x = 1 and the line y = 3 − x from x = 1 to x = 3, with the area under each piece shaded in a different color. 0 1 3 2 from 1 to 3 0 a 1

An integral with a variable endpoint of a function with absolute value, such as $H(x)=\int_0^x|f(t)|\,dt$, is written piecewise: on each piece you add what was accumulated before plus the new integral. Since $|f|$ is continuous, the fundamental theorem gives $H'(x)=|f(x)|$ for every $x$, even at a point where $f$ vanishes and $|f|$ has a "corner." For example, if $f(2)=0$, then $H$ is differentiable at $x=2$ and $H'(2)=|f(2)|=0$. What the theorem requires is that the integrand be continuous, not that it be differentiable.

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Given the function $f(x) = |x^2 - 2x|$, calculate the value of the definite integral on 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 on 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 on 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 on 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 whether the function is continuous on $x = 1$ 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
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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 local maximum of the function $f(x)$ on the interval $[0, 1]$.
  2. b) For the value of $a$ found, calculate using Barrow's Rule the area of the region enclosed between both graphs.
0 0.5 1 1 x y f(x) = sin(πx) g(x) = ax Maximum Area
Solve

Change of variable in the definite integral

If $u=g(x)$, then $du=g'(x)\,dx$. In the definite integral the limits change too.

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

Without doing 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:

$$\int_{-2}^{2} (x^5 + 3x^3 - x) \, dx$$

The figure shows the graph of $f(x)=x^5+3x^3-x$ on the interval $[-2,\,2]$, with the regions between the curve and the $X$-axis shaded.

Graph of f(x) = x⁵ + 3x³ − x between x = −2 and x = 2, with the regions between the curve and the X-axis shadedxy−220y = f(x)
Solve
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Let $f(x)$ be a continuous function on $\mathbb{R}$ 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 on $\mathbb{R}$ 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:

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

Use the substitution method (change of variables) in the definite integrals to relate them to the initial given value.

Solve

Average value of a function

The average of a function on an interval is the height of a rectangle that has the same area as the region under the curve.

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Find the mean value of the function $f(x) = \sqrt{x+1}$ on 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 $f(x) = \sqrt{x+1}$ on 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

Summary

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