Integral Calculus

Integral Calculus is the branch of calculus that deals with topics related to integration. Integrals are major components of calculus and are very useful in solving various problems based on real life. Some of such problems are the Basel problem, the problem of squaring the circle, the Gaussian integral, etc.

Fundamental Theorems of Integral Calculus

The integral represents the area under the curve. There are two fundamental theorems of integral calculus:

• First Fundamental Theorem of Integral Calculus
• Second Fundamental Theorem of Integral Calculus

First Fundamental Theorem of Integral Calculus

The first fundamental theorem of integral calculus states that if P(x) = ∫ f(x) dx is a continuous function on the interval [a, b], then P'(x) = f(x) for all x ∈ [a, b].

Second Fundamental Theorem of Integral Calculus

The second fundamental theorem of integral calculus states that if f(x) is a continuous function on the interval [a, b] and p(x) is the antiderivative of f(x), then \int\limits_a^b f(x) dx = p(b) - p(a) where the integral is called as a definite integral and a and b are called as lower and upper limits of the integral respectively.

Integral Definition

The antiderivative of the function f is called the integral of f. The reverse of differentiation is called as integration. The integral is also called as the primitive function of f or Newton-Leibnitz integral. The integral is the area under the curve.

Integrals can be classified as:

• Definite Integral: Integrals which are bounded by the limits
• Indefinite Integral: Integrals that do not have the limit of integration
• Improper Integrals: Integrals whose integrand is not bounded, or the limit of the integral is infinity

Multiple Integrals

Multiple integrals are integrals with more than one variable. There are two types of multiple integrals. They are:

• Double Integral
• Triple Integral

Double Integral

The integral of two variable functions under a specified region is called double integration. It is denoted by ∬.

Let's see the following example to learn more about the process of double integral.

Example: Evaluate the double integral ∬_R x² + y² dA, where R is the region bounded by the curves y = x² and y = 2x.

Answer:

The curves y = x² and y = 2x intersect at the point (0, 0) and (2, 4).

Thus, the limit of integration for variable x is: 0 ≤ x ≤ 2.

and for interval [0, 2], x² ≤ 2x.

∬_R x² + y² dA = \int_{0}^{2} \int_{x^2}^{2x} (x^2 + y^2) dy dx

Let I' = I' = \int_{x^2}^{2x} (x^2 + y^2) dy

\Rightarrow I' = \int_{x^2}^{2x} (x^2 + y^2) \, dy \\ \Rightarrow I' = \left[ x^2y + \frac{y^3}{3} \right]_{x^2}^{2x} \\ \Rightarrow I' = 2x^3 + \frac{8x^3}{3} - \frac{x^6}{3} + \frac{x^6}{3} \\ \Rightarrow I' = \frac{14x^3}{3}

Thus, ∬_R x² + y² dA = \int_{0}^{2} \frac{14x^3}{3} dx

⇒ ∬_R x² + y² dA = \frac{14}{3} \times \left[\frac{x^4}{4}\right]_{0}^{2}

⇒ ∬_R x² + y² dA = (14/3) × [(16/4) - 0]

⇒ ∬_R x² + y² dA = 56/3

Triple Integral

The integral of three variable functions under the 3-D region is called triple integration. It is denoted by ∭.

Let's consider an example to learn how to calculate the triple integral.

Certainly! Here's the direct solution using LaTeX code for the triple integral example:

Example: Evaluate the triple integral ∭_V xyz dV, where V is defined by 0 ≤ x ≤ 1, 0 ≤ y ≤ 2, and 0 ≤ z ≤ 3.

Solution:

Given the region V: 0 ≤ x ≤ 1, 0 ≤ y ≤ 2, and 0 ≤ z ≤ 3, the triple integral can be evaluated as follows:

\begin{aligned}\iiint V xyz \, dV &= \int_{0}^{1} \int_{0}^{2} \int_{0}^{3} xyz \, dz \, dy \, dx \\\Rightarrow \iiint V xyz &= \int_{0}^{1} \int_{0}^{2} \left[ \frac{xy z^2}{2} \right]_{0}^{3} \, dy \, dx \\\Rightarrow \iiint V xyz &= \int_{0}^{1} \int_{0}^{2} \frac{27xy}{2} \, dy \, dx \\\Rightarrow \iiint V xyz &= \int_{0}^{1} \left[ \frac{27xy^2}{4} \right]_{0}^{2} \, dx \\\Rightarrow \iiint V xyz &= \int_{0}^{1} 27x \, dx \\\Rightarrow \iiint V xyz &= \left[ \frac{27x^2}{2} \right]_{0}^{1} \\\Rightarrow \iiint V xyz &= \frac{27}{2}\end{aligned}

So, the value of the triple integral ∭_V xyz dV over the region V is 27/2.

Methods to Find Integrals

There are multiple types of integrals which can be solved using different methods. Some integrals can be directly solved by applying formulas. To solve some integrals, we use the following methods:

• Integration by Substitution
• Integration by Parts
• Integration by Partial Fraction

Integration by Substitution

To understand the method of Integration by Substitution, we can see the following example or we can explore the article mentioned above.

Example: Evaluate: ∫[(2x)/ {5x² +1}]dx

Solution:

Let 5x² + 1 = t. Then, d(5x² + 1) = dt ⇒ 10 x dx = dt

⇒ ∫[(2x)/ {5x² +1}]dx = ∫[2x / (t×10x)]dt

⇒ ∫[(2x)/ {5x² +1}]dx = ∫[1 / (5t)]dt

⇒ ∫[(2x)/ {5x² +1}]dx = (1/5) ∫[1 / t]dt

⇒ ∫[(2x)/ {5x² +1}]dx = [(log t) / 5] + C

Integration by Parts

Let's consider an example for better understanding.

Example: Evaluate the integral ∫e^x x dx

Solution:

Let I = ∫e^x x dx

This integral can be solved by integration by parts (ILATE rule)

According to ILATE rule the first function u = x (algebraic), v = e^x (exponent)

The formula of integration by parts

⇒ ∫u.v dx = u∫v dx - ∫[(du/dx)∫v dx] dx

⇒ ∫x e^x dx = x∫e^x dx - ∫[(dx/dx)∫e^x dx] dx

⇒ ∫x e^x dx = x e^x - ∫[(1)e^x] dx + C1

⇒ ∫x e^x dx = x e^x - ∫e^x dx + C1

⇒ ∫x e^x dx = x e^x - e^x + C1 + C2

⇒ ∫x e^x dx = e^x(x - 1) + C [C = C1 + C2]

Integration by Partial Fraction

For better understanding, let's consider the following example.

Example: Evaluate the integral ∫[x/ {(x - 1)(x - 2)}]dx

Solution:

I = ∫[x/ {(x - 1)(x - 2)}]dx

These types of integrals can be solved using partial fraction method.

[x/ {(x - 1)(x - 2)}] = A/(x - 1) + B /(x - 2)

⇒ [x/ {(x - 1)(x - 2)}] = [A(x-2) + B(x-1)]/[(x - 1) (x - 2)]

Equating the numerators

x = A(x-2) + B(x - 1)

⇒ x = Ax + Bx -2A - B

⇒ x = (A + B)x - (2A + B)

Comparing coefficients

A + B = 1 . . . (1)

-2A - B = 0 . . . (2)

From (1) and (2)

B = -2A

Thus, A +(-2A) = 1

⇒ -A = 1

⇒ A = -1

Thus, B = 2

Putting values of A and B in (I)

[x/ {(x - 1)(x - 2)}] = (-1)/(x - 1) + 2 /(x - 2)

⇒ ∫[x/ {(x - 1)(x - 2)}] = ∫[(-1)/(x - 1) + 2 /(x - 2)] dx

⇒ ∫[x/ {(x - 1)(x - 2)}] = ∫[(-1)/(x - 1)]dx + ∫[2 /(x - 2)] dx

⇒ ∫[x/ {(x - 1)(x - 2)}] = -∫[1/(x - 1)]dx + 2∫[1 /(x - 2)] dx

⇒ ∫[x/ {(x - 1)(x - 2)}] = -ln(x - 1) + 2 ln(x - 2) + C1 + C2

⇒ ∫[x/ {(x - 1)(x - 2)}] = -ln(x - 1) + 2 ln(x - 2) + C [C=C1 + C2]

Applications of Integral Calculus

Integral calculus has different applications. Some of them are:

• To find the area under the curve.
• To find the area between two curves.
• To find the volumes.
• To find displacement and motion.

Example 1: Find the area of the region bounded by the curve y² = 16x, x = 1, x=3 and the x-axis in first quadrant.

Solution:

First, we draw the figure of the curve y² = 16x, x = 1, x=3 and the x-axis in first quadrant

y² = 16x

⇒ y = ±√(16x)

⇒ y = ±4√x . . . (1)

We will take positive value of y as we have to find area under the first quadrant.

The required area is the shaded region PQRS.

Required area = \int\limits^3_1 y dx

⇒ Required area = \int\limits^3_1 4√xdx [From equation 1]

⇒ Required area = 4[x^½+1 / {(1/2)+1}]₁³

⇒ Required area = 4[x^3/2 / (3/2)]₁³

⇒ Required area = 4[(2x^3/2)/3]₁³

⇒ Required area = (8/3) [ 3^3/2 - 1^3/2]

⇒ Required area = (8/3) [3√3 - 1]

⇒ Required area = [8√3 - (8/3)] sq units

Example 2: Find the volume of solid bounded between the region x² +y² ≤ 4 and 0 ≤ z ≤ 1.

Solution:

From the given equation radius of disc = 2 units.

Required Volume = ∭_V dx dy dz

⇒ Required volume = ∬_R\int\limits_{z=0}^{z=1}(dz)dx dy

⇒ Required volume = ∬_R[z]₀¹dxdy = ∬_R[1 - 0]dx dy

⇒ Required volume = ∬_R[z]₀¹dxdy = ∬_R (1)dx dy

⇒ Required volume = ∬_R dx dy

Since, ∬_Rdx dy = Area (given shape is a disc which is a circle and area = πr² )

⇒ Required volume = π(2)²

⇒ Required volume = 4π cubic units

Example 3: Find the displacement of the particle over the interval [1, 3] if the velocity of the particle is given by v(t) = 3t² + 2t.

Solution:

Given the velocity v(t) = 3t² + 2t and the interval [1, 3]

Displacement S(t) = \int\limits_a^bv(t) dt

Here, a = 1 and b = 3 by the given interval

⇒ S(t) = \int\limits_1^3(3t² + 2t)dt

⇒ S(t) = \int\limits_1^33t²dt + \int\limits_1^32tdt

⇒ S(t) = 3[(t³/3)]₁³ + 2[(t²/ 2)]₁³

⇒ S(t) = [t³]₁³ + [t²]₁³

⇒ S(t) = [3³ - 1³] + [3² - 1²]

⇒ S(t) = [27 - 1] + [ 9 - 1]

⇒ S(t) = 26 + 8

⇒ S(t) = 34 units

The displacement is 34 units.

Differential vs Integral Calculus

The key differences between Differential calculus and Integral calculus are listed in the following table.

Integral Calculus Examples

Example 1: Solve: ∫x⁹dx

Answer:

∫x⁹dx = [x^{9 + 1} / (9 + 1)] + C

⇒ ∫x⁹dx = (x¹⁰ / 10) + C

Example 2: Solve ∫ 2^{x + 3}dx.

Answer:

∫ 2^{x + 3}dx = ∫ 2^x 2³dx

⇒ ∫ 2^{x + 3}dx = 8∫ 2^x dx

⇒ ∫ 2^{x + 3}dx = 8[2^x / log_e2] + C

Example 3: Evaluate: ∫(x³ + 3x² + 5x + 6)dx

Answer:

∫(x³ + 3x² + 5x + 6) dx = ∫x³dx + ∫ 3x²dx + ∫ 5x dx + ∫ 6dx

⇒ ∫(x³ + 3x² + 5x + 6) dx = (x³⁺¹ / (3+1)) + 3[(x²⁺¹ / (2+1))] + 5[(x¹⁺¹ / (1+1))] + 6[x⁰⁺¹/ (0+1)] + C

⇒ ∫(x³ + 3x² + 5x + 6) dx = (x⁴ / 4) + x³ + (5/2)x² + 6x + C

Example 4: Solve ∫(x + 6cos x) dx.

Answer:

∫(x + 6cos x) dx = ∫x dx + ∫cos x dx

⇒ ∫(x + 6cos x) dx = ∫x dx + 6 ∫cos x dx

⇒ ∫(x + 6cos x) dx = [x¹⁺¹ / (1+1)] + 6sin x + C

⇒ ∫(x + 6cos x) dx = [x² / 2] + 6sin x + C

Example 5: Evaluate ∫[1/{4√(x² - 16)}].

Answer:

∫[1/{4√(x² - 16)}] = ∫[1/{4√(x² - 4²)}]

⇒ ∫ [1 / {x√(x² - a²)}] dx = (1/a) sec^-1(x/a) + C

⇒ ∫[1/{4√(x² - 16)}] = (1/4) sec^-1(x/4) + C

Practice Problems on Integral Calculus

1. Calculate the integral of the function: ∫(3x² - 5x + 4) dx

2. Evaluate the integral involving a basic trigonometric function: ∫cos⁡(2x) dx

3. Use substitution to solve the integral: ∫x sin(x²) dx

4. Calculate the integral using the method of integration by parts: ∫x² e^x dx

5. Find the area under the curve from x = 0 to x = ? for the function: ∫₀^pi Sin(x) dx