Question

$$ \text { use integration by parts to evaluate each integral. } $$ $$ \int t \sqrt{t+1} d t $$

Short answer

The integral evaluates to \( \frac{2}{3} t (t+1)^{3/2} - \frac{4}{15} (t+1)^{5/2} + C \).

Step-by-step solution

Identify Parts for Integration by Parts

To apply integration by parts, we use the formula \( \int u \, dv = uv - \int v \, du \). We need to identify \( u \) and \( dv \) from \( \int t \sqrt{t+1} \, dt \). Let \( u = t \) and differentiate it to get \( du = dt \). Let \( dv = \sqrt{t+1} \, dt \).

Integrate dv to find v

Integrate \( dv = \sqrt{t+1} \, dt \) to find \( v \). Let \( x = t+1 \), then \( \sqrt{x} = x^{1/2} \) and \( dx = dt \). Integrating, \( \int \sqrt{x} \, dx = \int x^{1/2} \, dx = \frac{2}{3} x^{3/2} + C \). Substitute back to get \( v = \frac{2}{3} (t+1)^{3/2} \).

Apply Integration by Parts Formula

Substitute the expressions into the integration by parts formula. We have \( u = t \), \( du = dt \), \( v = \frac{2}{3} (t+1)^{3/2} \), and \( \int v \, du = \int \frac{2}{3} (t+1)^{3/2} \, dt \). This gives: \( \int t \sqrt{t+1} \, dt = \frac{2}{3} t (t+1)^{3/2} - \int \frac{2}{3} (t+1)^{3/2} \, dt \).

Integrate Remaining Integral

Now integrate \( \int \frac{2}{3} (t+1)^{3/2} \, dt \) by substituting back. This becomes \( \int \frac{2}{3} (t+1)^{3/2} \, dt = \frac{2}{3} \cdot \frac{2}{5} (t+1)^{5/2} = \frac{4}{15} (t+1)^{5/2} \).

Simplify and Combine the Expressions

Combine the terms from Step 3 by substituting back the integrated term from Step 4. This results in \( \frac{2}{3} t (t+1)^{3/2} - \frac{4}{15} (t+1)^{5/2} + C \), where \( C \) is the constant of integration.

Key concepts

Definite Integrals

A definite integral is a fundamental concept in calculus that represents the area under the curve of a function over a specified interval. It is denoted by \[ \int_a^b f(x) \, dx \]where \(a\) and \(b\) are the limits of integration and \(f(x)\) is the function being integrated.

Definite integrals are important because they allow us to find the total accumulation of quantity, such as area, volume, or other physical quantities, across an interval. Unlike indefinite integrals, definite integrals yield a number as a result rather than a function.

• They provide the accumulated value.
• Boundaries \(a\) and \(b\) are given.
• Result is a specific number, not a function.

The process involves finding an antiderivative of the function, evaluating this antiderivative at the upper and lower bounds, and subtracting the results.

Antiderivatives

Antiderivatives, also known as indefinite integrals, are the opposite of derivatives. They involve finding a function whose derivative results in the original function given. An antiderivative of a function \(f(x)\) is denoted by\[ \int f(x) \, dx = F(x) + C \]where \(F(x)\) is an antiderivative of \(f(x)\) and \(C\) is a constant of integration.

Antiderivatives are crucial when dealing with integration because they help calculate areas under curves, reverse the process of differentiation, and solve differential equations.

• Basic tool for finding integrals.
• Essential for solving definite integrals (with specific limits).
• Include a constant \(C\) as solutions are not unique.

In the given example, finding antiderivatives involves working through steps of integration by parts, particularly for complex expressions involving multiple terms.

Calculus Integration Techniques

Calculus offers various integration techniques to find antiderivatives and definite integrals for more complex functions. One fundamental method is "integration by parts," which is helpful for products of functions like polynomials and exponentials.

The integration by parts formula is given by:

• \[ \int u \, dv = uv - \int v \, du \]

where \(u\) and \(dv\) are parts of the original integral, \(du\) is the derivative of \(u\), and \(v\) is the antiderivative of \(dv\).

Other techniques include:

• Substitution: Simplifies integrals by changing variables.
• Partial Fractions: Breaks down rational functions into simpler parts.
• Trigonometric Integrals: Handles integrals involving trigonometric functions.

These techniques allow us to tackle a wide range of integrals efficiently.

Integral Calculus Steps

Integral calculus involves a series of systematic steps to evaluate integrals effectively. These steps are designed to decompose complex problems into manageable tasks, leading to accurate solutions.

For integration by parts, specifically, the method is broken down into distinct parts:

• Identify \(u\) and \(dv\) from the integral.
• Differentiate \(u\) to get \(du\), and integrate \(dv\) to find \(v\).
• Apply the integration by parts formula: \( \int u \, dv = uv - \int v \, du \).
• Solve the resulting integral and combine terms.

These steps may involve substitution and multiple iterations, especially for complex integrals, ensuring each stage naturally leads to the next. The objective is to simplify the integral iteratively until it reaches a solvable form, as demonstrated in the original exercise using integration by parts.