This article will delve into the fascinating world of **u-substitution** in **definite integrals**, aiming to provide readers with a comprehensive understanding of its concept, application, and significance. We’ll unravel its intricacies, explore its properties, and demonstrate its utility with **practical examples**, offering a holistic view of this vital **calculus** tool.

## Definition of U Substitution Definite Integral

In **calculus**, **u-substitution** is a method for finding integrals. In u-substitution, the substitution **u = g(x)** is made to simplify the integral. When a **definite integral** is considered, the limits of the integral are also changed according to the new variable ‘**u**.’

More formally, if you have an **integral** of form **∫f(g(x)) * g'(x) dx**, you can make a **substitution** to simplify this to **∫f(u) du**, where **u** is a function **u = g(x)**. The corresponding limits of the integral in terms of ‘**u**‘ are found by substituting the original ‘**x**‘ limits into the function **u = g(x)**.

**U-substitution**, essentially the reverse process of the chain rule of differentiation, can greatly simplify finding many** integrals**.

**Example**

∫x² √(x³ + 1) dx; **[0 to 2]**

Figure-1.

### Solution

Let **u = x³ + 1 du = 3x² dx**

Substitute the limits: When x = 0, u = 0³ + 1 = 1 When x = 2, u = 2³ + 1 = 9

The integral becomes:

∫(1/3)√u du, [1 to 9]

Applying power rule and u-substitution:

= (1/3) * (2/3) * (u³∕²)) evaluated from 1 to 9

= (2/9) * (9√9 – 1√1)

= (2/9) * (27 – 1)

= (2/9) * 26

= 52/9

Therefore, ∫[0 to 2] x² √(x³ + 1) dx = **52/9**

**Evaluation Process**

The **evaluation process** of **u-substitution** in **definite integrals** involves several steps, as outlined below:

**Identify a Substitution**

Start by identifying a part of the** integral** that could simplify the problem if substituted with a single variable, ‘**u**.’ Typically, you’d select a function that makes the integral look simpler when** substituted** or a function whose **derivative** is present elsewhere in the** integral**.

**Make the Substitution**

Replace the chosen part of the function with ‘**u**‘. So, if you have a function of the form **∫f(g(x)) * g'(x) dx**, you substitute **u = g(x)**, so the integral becomes **∫f(u) * du**.

**Change the Limits of Integration**

For **definite integrals**, remember to change the limits of integration. If the original limits of the **x-integral** are **a** and **b**, then substitute these into your equation **u = g(x)** to find the new limits for **u**. Let’s say these are **c** and **d**.

**Perform the Integral with the New Variable**

With a **simpler function** and **limits**, perform the integration in terms of ‘**u**‘. This will yield a new function, let’s call it **F(u)**.

**Substitute ‘u’ Back In**

Replace ‘**u**‘ with the original function **g(x)** in the **antiderivative**. Now we have a new function **F(g(x))**.

**Evaluate Between the New Limits**

Finally, **substitute** the new limits (in terms of ‘**u**‘) into the **antiderivative**, calculate the **difference,** and get the final result. That is, you’ll be finding **F(d) – F(c)**.

**Exercise **

**Example 1**

∫(3x² + 2x + 1) $e^{(x³ + x² + x)}$ dx; **[-1 to 1]**

### Solution

Let **u = x³ + x² + x du = (3x² + 2x + 1) dx**

Substitute the limits: When x = -1, u = (-1)³ + (-1)² + (-1) = -1 When x = 1, u = 1³ + 1² + 1 = 3

The integral becomes:

∫eᵘ du; [-1 to 3]

Applying the power rule and u-substitution:

= eᵘ evaluated from -1 to 3 = e³ – e⁻¹

Therefore:

∫(3x² + 2x + 1) $e^{(x³ + x² + x)}$ dx; [-1 to 1]

= e³ – e⁻¹

**Example 2**

∫x³ √(x⁴ – 1) dx; **[1 to 2] **

### Solution

Let **u = x⁴ – 1 du = 4x³ dx**

Substitute the limits: When x = 1, u = 1⁴ – 1 = 0 When x = 2, u = 2⁴ – 1 = 15

The integral becomes:

∫(1/4) √u du; [0 to 15]

Applying power rule and u-substitution:

= (1/4) * (2/3) * (u³∕²) evaluated from 0 to 15

= (1/4) * (2/3) * (15³∕² – 0³∕²)

= (1/4) * (2/3) * (15³∕²)

= (1/6) * (15³∕²)

Therefore:

∫x³ √(x⁴ – 1) dx; [1 to 2]

= (1/6) * (15³∕²)

**Example 3**

∫sin(2θ) cos²(θ) dθ; **[-π/2 to π/2] **

### Solution

Let **u = cos(θ) du = -sin(θ) dθ**

Substitute the limits: When θ = -π/2, u = cos(-π/2) = 0 When θ = π/2, u = cos(π/2) = 0

The integral becomes:

∫-u² du; [0 to 0]

Since the limits are the same, the integral evaluates to 0.

Therefore:

∫sin(2θ) cos²(θ) dθ; [-π/2 to π/2]

= 0

**Example 4**

∫(x² – 2x + 1) √(1 – x²) dx;** [-1 to 1] **

Figure-2.

### Solution

Let **u = 1 – x² du = -2x dx**

Substitute the limits: When x = -1, u = 1 – (-1)² = 0 When x = 1, u = 1 – 1² = 0

The integral becomes:

∫-(1/2) √u du; [0 to 0]

Since the limits are the same, the integral evaluates to 0.

Therefore:

∫(x² – 2x + 1) √(1 – x²) dx; [-1 to 1]

= 0

**Example 5**

∫x³ $e^{(x⁴)}$ dx; **[0 to 1] **

### Solution

Let **u = x⁴ du = 4x³ dx**

Substitute the limits: When x = 0, u = 0⁴ = 0 When x = 1, u = 1⁴ = 1

The integral becomes:

∫(1/4) eᵘ du; [0 to 1]

= (1/4) * ∫eᵘ du; [0 to 1]

= (1/4) * (e¹ – e⁰)

= (1/4) * (e – 1)

Therefore:

∫x³ $e^{(x⁴)}$ dx = (1/4) * (e – 1); [0 to 1]

**Example 6**

∫sin³(θ) cos²(θ) dθ; **[-π/2 to π/2] **

Figure-3.

### Solution

Let **u = cos(θ) du = -sin(θ) dθ**

Substitute the limits: When θ = -π/2, u = cos(-π/2) = 0 When θ = π/2, u = cos(π/2) = 0

The integral becomes:

∫-u² (1 – u²) du; [0 to 0]

Since the limits are the same, the integral evaluates to 0.

Therefore:

∫sin³(θ) cos²(θ) dθ = 0; [-π/2 to π/2]

**Applications **

The concept of** u-substitution in definite integrals** is fundamental to **calculus** and thus finds extensive applications across multiple disciplines that use **calculus** in their work. Here are a few of those applications:

**Physics**

In **physics**, integration, including **u-substitution**, is used to calculate quantities like work done by a variable force, electric and magnetic fields created by charge and current distributions, or the **moment of inertia** of an** object** with a **complex shape**.

**Engineering**

In many **engineering** problems, especially those involving **calculus of variations**, **u-substitution** simplifies the integrals. It is frequently used in **electrical engineering**, where integration is used to calculate quantities like charge, energy, power, etc., given their rates.

**Economics**

In **economics**, integration is used in numerous ways, such as determining **consumer** and **producer surplus**, calculating the **present value** of a continuous income stream, or modeling and solving **dynamic equilibrium** problems. The method of **u-substitution** often simplifies these calculations.

**Statistics and Probability**

**U-substitution** is often used for **probability density functions**, especially **continuous random variables**. It is also used in the process of **normalization**, where a probability density function is made to integrate to 1.

**Biology**

In **biology**, integrals, including those simplified by **u-substitution**, are used in growth and decay models, **population dynamics**, and in interpreting the behavior of systems over continuous intervals.

**Computer Graphics**

In the field of **computer graphics**, and particularly in rendering and animation, integrals are used to calculate light and color values in a scene. **U-substitution** is often used to simplify these integrals, making them computationally more efficient.

**Medicine**

In **biomedical engineering**, the **u-substitution** method is often used in signal and image processing applications, such as modeling the response of a biological system to a drug dosage over time.

**Environmental Sciences**

In studying **pollutant spread** or **population dynamics** of certain species, the **u-substitution** method in definite integrals can be employed to model and predict behaviors over time.

**Chemistry**

In **physical chemistry**, integration using **u-substitution** is used for solving **differential equations** related to reaction rates. It’s also used in **quantum mechanics** to calculate probabilities from wave functions.

**Geography and Meteorology**

**U-substitution** in integrals can be used in models predicting weather patterns and climate change, as these often involve calculations of accumulated changes over time or space.

**Astronomy and Space Science**

Integration calculates various physical quantities, such as **gravitational** and **electromagnetic fields**, often involving complex or spherical coordinates where **u-substitution** can simplify the integrals.

**Operations Research**

This field often requires the **optimization** of certain **resources**. The associated problems frequently involve **integration**, where **u-substitution** can be used to simplify complex relationships.

**Machine Learning and Data Science**

Integration is fundamental to **machine learning** and **data science** aspects, like calculating areas under the **ROC curve**, probability densities, and more. **U-substitution** is a helpful tool in solving these integrals.

**Psychophysics**

In the field of **psychophysics**, which investigates the relationship between stimuli (which are **physical**) and the sensations and perceptions they affect (which are **psychological**), definite integrals using **u-substitution** are often used to quantify the relationship between the physical stimulus and the perceived sensation.

**Finance and Actuarial Science**

**Integration** techniques, including **u-substitution**, are used in calculating the present and future values of **continuous income streams**, **pricing complex financial derivatives**, and **building models** in **actuarial science**.

*All images were created with GeoGebra and MATLAB.*