Indefinite Integral

FIRST FUNDAMENTAL THEOREM OF CALCULUS

Definition: The First Fundamental Theorem of Calculus relates the concept of an antiderivative with the concept of definite integrals.
Statement: If f is a continuous function on the interval [a, b], and F is an antiderivative of f on [a, b], then
$$ \int_{a}^{b} f(x) ,dx = F(b) - F(a) $$
It provides a method to evaluate definite integrals by finding antiderivatives.
The First Fundamental Theorem of Calculus establishes a connection between differentiation and integration.
It enables us to evaluate integrals using techniques of differentiation.

Example 1:

Let f(x) = 3x^2 - 6x. Find the value of ∫[1, 3] f(x) dx using the First Fundamental Theorem of Calculus. Solution:

Step 1: Find an antiderivative of f(x).
- By using the power rule of differentiation, an antiderivative of f(x) is F(x) = x^3 - 3x^2 + C.
Step 2: Evaluate F(3) and F(1).
- F(3) = 3^3 - 3(3)^2 + C = 27 - 27 + C = C
- F(1) = 1^3 - 3(1)^2 + C = 1 - 3 + C = C - 2
Step 3: Compute the definite integral using the First Fundamental Theorem of Calculus.
- ∫[1, 3] f(x) dx = F(3) - F(1) = C - (C - 2) = 2

Example 2:

Evaluate the definite integral ∫[-1, 1] (x^2 + 4x - 6) dx. Solution:

Step 1: Find an antiderivative of the integrand.
- An antiderivative of f(x) = x^2 + 4x - 6 is F(x) = (1/3)x^3 + 2x^2 - 6x + C.
Step 2: Evaluate F(1) and F(-1).
- F(1) = (1/3)(1)^3 + 2(1)^2 - 6(1) + C = (1/3) + 2 - 6 + C = C - (50/3)
- F(-1) = (1/3)(-1)^3 + 2(-1)^2 - 6(-1) + C = -(1/3) + 2 + 6 + C = C + (19/3)
Step 3: Compute the definite integral.
- ∫[-1, 1] (x^2 + 4x - 6) dx = F(1) - F(-1) = (C - (50/3)) - (C + (19/3)) = -69/3

Advantages of the First Fundamental Theorem of Calculus:

It provides a simple method to evaluate definite integrals.
It relates differentiation and integration, connecting two important concepts in calculus.
It can be used to find functions when only information about their derivatives is given.
It allows for the calculation of areas between curves using definite integrals.

Summary:

The First Fundamental Theorem of Calculus relates the concept of antiderivatives with definite integrals.
It states that if f is continuous on [a, b] and F is an antiderivative of f on [a, b], then ∫[a, b] f(x) dx = F(b) - F(a).
This theorem enables us to evaluate definite integrals using techniques of differentiation.
It is a fundamental result that connects differentiation and integration, and has various applications in calculus.

Integrals of Elementary Functions

Many elementary functions have well-known antiderivatives that can be found using techniques of integration.
Some common integrals include:
- ∫ x^n dx = (1/(n+1))x^(n+1) + C, where n ≠ -1
- ∫ e^x dx = e^x + C
- ∫ sin(x) dx = -cos(x) + C
- ∫ cos(x) dx = sin(x) + C
- ∫ sec^2(x) dx = tan(x) + C
- ∫ sec(x)tan(x) dx = sec(x) + C
- ∫ 1/(x^2 + a^2) dx = (1/a) arctan(x/a) + C

Example:

Evaluate the integral ∫ x^3 + 2x^2 - 6x + 1 dx. Solution:

Using the power rule of integration, we have:
- ∫ x^3 + 2x^2 - 6x + 1 dx = (1/4)x^4 + (2/3)x^3 - (3/2)x^2 + x + C

Integration by Parts

Integration by parts is a technique used to evaluate integrals of products of functions.
It is based on the product rule of differentiation.
The formula for integration by parts is given by:
- ∫ u dv = uv - ∫ v du
This formula allows us to relate the integral of a product to the integral of the derivative of one of the functions.

Example:

Evaluate the integral ∫ x sin(x) dx. Solution:

Let u = x and dv = sin(x) dx.
Then du = dx and v = -cos(x).
Applying the formula for integration by parts, we have:
- ∫ x sin(x) dx = -x cos(x) - ∫ (-cos(x)) dx = -x cos(x) + sin(x) + C

Trigonometric Substitution

Trigonometric substitution is a technique used to evaluate integrals involving radical expressions and trigonometric functions.
It involves making a substitution using trigonometric identities to simplify the integral.
There are three common trigonometric substitutions:
- For integrals involving √(a^2 - x^2), we use x = a sinθ.
- For integrals involving √(a^2 + x^2), we use x = a tanθ.
- For integrals involving √(x^2 - a^2), we use x = a secθ.

Example:

Evaluate the integral ∫ (4x^2 + 1) / √(x^2 + 9) dx. Solution:

Let x = 3 tanθ.
Then dx = 3 sec^2θ dθ.
Substituting into the integral and simplifying, we have:
- ∫ (4(3 tanθ)^2 + 1) / √((3 tanθ)^2 + 9) (3 sec^2θ) dθ
- = ∫ (36 tan^2θ + 1) / √(9 tan^2θ + 9) sec^2θ dθ

Partial Fraction Decomposition

Partial fraction decomposition is a technique used to simplify and evaluate integrals that involve rational functions.
It involves breaking down a rational function into simpler fractions to make integration easier.
The basic idea is to write the rational function as a sum of simpler fractions with denominators that are powers of linear factors or irreducible quadratic factors.
This allows us to use known integral formulas to evaluate the integral.

Example:

Evaluate the integral ∫ (3x^2 - 5) / (x^3 + x^2 - 2x) dx. Solution:

First, we factor the denominator: x^3 + x^2 - 2x = x(x - 1)(x + 2).
Next, we express the rational function as a sum of partial fractions:
- (3x^2 - 5) / (x^3 + x^2 - 2x) = A/x + B/(x - 1) + C/(x + 2).
Now, we find the values of A, B, and C by equating numerators:
- 3x^2 - 5 = A(x - 1)(x + 2) + Bx(x + 2) + Cx(x - 1).

</section>
<section style="font-size: 50px";>
<h2 id="indefinite-integral---first-fundamental-theorem-of-calculus">Indefinite Integral - FIRST FUNDAMENTAL THEOREM OF CALCULUS</h2>
<ul>
<li>Definition: The First Fundamental Theorem of Calculus relates the concept of an antiderivative with the concept of definite integrals.</li>
<li>Statement: If f is a continuous function on the interval [a, b], and F is an antiderivative of f on [a, b], then<br>
$$ \int_{a}^{b} f(x) ,dx = F(b) - F(a) $$</li>
<li>It provides a method to evaluate definite integrals by finding antiderivatives.</li>
<li>The First Fundamental Theorem of Calculus establishes a connection between differentiation and integration.</li>
<li>It enables us to evaluate integrals using techniques of differentiation.</li>
</ul>
</section>
<section style="font-size: 50px";>
<h2 id="example-1-1">Example 1:</h2>
<p>Let f(x) = 3x^2 - 6x.
Find the value of ∫[1, 3] f(x) dx using the First Fundamental Theorem of Calculus.
Solution:</p>
<ul>
<li>Step 1: Find an antiderivative of f(x).</li>
<li>By using the power rule of differentiation, an antiderivative of f(x) is F(x) = x^3 - 3x^2 + C.</li>
<li>Step 2: Evaluate F(3) and F(1).</li>
<li>F(3) = 3^3 - 3(3)^2 + C = 27 - 27 + C = C</li>
<li>F(1) = 1^3 - 3(1)^2 + C = 1 - 3 + C = C - 2</li>
<li>Step 3: Compute the definite integral using the First Fundamental Theorem of Calculus.</li>
<li>∫[1, 3] f(x) dx = F(3) - F(1) = C - (C - 2) = 2</li>
</ul>
</section>
<section style="font-size: 50px";>
<h2 id="example-2-1">Example 2:</h2>
<p>Evaluate the definite integral ∫[-1, 1] (x^2 + 4x - 6) dx.
Solution:</p>
<ul>
<li>Step 1: Find an antiderivative of the integrand.</li>
<li>An antiderivative of f(x) = x^2 + 4x - 6 is F(x) = (1/3)x^3 + 2x^2 - 6x + C.</li>
<li>Step 2: Evaluate F(1) and F(-1).</li>
<li>F(1) = (1/3)(1)^3 + 2(1)^2 - 6(1) + C = (1/3) + 2 - 6 + C = C - (50/3)</li>
<li>F(-1) = (1/3)(-1)^3 + 2(-1)^2 - 6(-1) + C = -(1/3) + 2 + 6 + C = C + (19/3)</li>
<li>Step 3: Compute the definite integral.</li>
<li>∫[-1, 1] (x^2 + 4x - 6) dx = F(1) - F(-1) = (C - (50/3)) - (C + (19/3)) = -69/3</li>
</ul>
</section>
<section style="font-size: 50px";>
<h2 id="advantages-of-the-first-fundamental-theorem-of-calculus-1">Advantages of the First Fundamental Theorem of Calculus:</h2>
<ul>
<li>It provides a simple method to evaluate definite integrals.</li>
<li>It relates differentiation and integration, connecting two important concepts in calculus.</li>
<li>It can be used to find functions when only information about their derivatives is given.</li>
<li>It allows for the calculation of areas between curves using definite integrals.</li>
</ul>
</section>
<section style="font-size: 50px";>
<h2 id="summary-1">Summary:</h2>
<ul>
<li>The First Fundamental Theorem of Calculus relates the concept of antiderivatives with definite integrals.</li>
<li>It states that if f is continuous on [a, b] and F is an antiderivative of f on [a, b], then ∫[a, b] f(x) dx = F(b) - F(a).</li>
<li>This theorem enables us to evaluate definite integrals using techniques of differentiation.</li>
<li>It is a fundamental result that connects differentiation and integration, and has various applications in calculus.</li>
</ul>
</section>

</div>
				</div>

function transformCallouts() {
      var slides = document.querySelectorAll('.reveal .slides section');
      slides.forEach(function (slide) {
        var data = document.getElementsByTagName("Success");
        for (let i = 0; i < data.length; i++) {
          data[i].classList.add("callout");
          data[i].classList.add("callout-success");
        }

var data = document.getElementsByTagName("Default");
        for (let i = 0; i < data.length; i++) {
          data[i].classList.add("callout");
          data[i].classList.add("callout-default");
        }

var data = document.getElementsByTagName("Warning");
        for (let i = 0; i < data.length; i++) {
          data[i].classList.add("callout");
          data[i].classList.add("callout-warning");
        }

var data = document.getElementsByTagName("Primary");
        for (let i = 0; i < data.length; i++) {
          data[i].classList.add("callout");
          data[i].classList.add("callout-primary");
        }

var data = document.getElementsByTagName("Danger");
        for (let i = 0; i < data.length; i++) {
          data[i].classList.add("callout");
          data[i].classList.add("callout-danger");
        }
      });
    };

function addrfit() {
      var slides = document.querySelectorAll('.reveal');
      slides.forEach(function (slide) {
        var temp = document.getElementsByTagName("h2");

if (temp.length > 0) {
          for (let i = 0; i < temp.length; i++) {
            temp[i].classList.add("r-fit-text");
          }
        }

var temp3 = document.getElementsByTagName("h1");

if (temp3.length > 0) {
          for (let i = 0; i < temp3.length; i++) {
            temp3[i].classList.add("r-fit-text");
          }
        }

var temp6 = document.getElementsByClassName("text-container");

if (temp6.length > 0) {
          for (let i = 0; i < temp6.length; i++) {
            var charcount = 0;
            var ptags = temp6[i].getElementsByTagName("p");
            try {
              for (let j = 0; j < ptags.length; j++) {
                temp = ptags[j].outerText.replaceAll(" ", "");
                temp = temp.replaceAll("\n", "");
                charcount += temp.length;
              }
            } catch (err) {
              console.log(err);
            }

var font = 35;

if (charcount > 50 && charcount < 100) {
              font = 35;
            } else if (charcount > 100 && charcount < 120) {
              font = 33;
            } else if (charcount > 120 && charcount < 150) {
              font = 33;
            } else if (charcount > 150 && charcount < 170) {
              font = 31;
            } else if (charcount > 170 && charcount < 200) {
              font = 30;
            } else if (charcount > 200 && charcount < 230) {
              font = 29;
            } else if (charcount > 230 && charcount < 250) {
              font = 28;
            } else if (charcount > 250 && charcount < 270) {
              font = 27;
            } else if (charcount > 270 && charcount < 300) {
              font = 25;
            } else if (charcount > 300 && charcount < 330) {
              font = 23;
            } else if (charcount > 330 && charcount < 360) {
              font = 24;
            } else if (charcount > 360 && charcount < 390) {
              font = 22;
            } else if (charcount > 390 && charcount < 420) {
              font = 21;
            } else if (charcount > 420) {
              font = 20;
            }

temp6[i].style.fontSize = `${font}px`;
          }
        }
      })
    }

Reveal.addEventListener('ready', addrfit);
    Reveal.addEventListener('ready', transformCallouts);
  </script>