Problem 76 - Applied Calculus

Question

State (without carrying them out) two different methods to find \(\int t e^{t} d t\)

Step-by-Step Solution

Answer

Use integration by parts or tabular integration.

Step 1: Identify the Integration by Parts Formula

Integration by parts is a common technique for integrating products of functions. The formula is given by: \( \int u \, dv = uv - \int v \, du \). Identify parts of the integrand to use: let \( u = t \) and \( dv = e^t \, dt \).

Step 2: Differentiate and Integrate Parts

Find \( du \) by differentiating \( u \), so \( du = dt \). Integrate \( dv \) to find \( v \), so \( v = e^t \).

Step 3: Method 1 - Apply Integration by Parts

Apply the integration by parts formula: \( \int t e^t \, dt = te^t - \int e^t \, dt \). Now integrate \( \int e^t \, dt \) to get \( e^t \). Substitute back: \( te^t - e^t + C \), where \( C \) is a constant of integration.

Step 4: Method 2 - Use Tabular Integration

Another way is tabular integration, which simplifies integration by parts for repeated use. Align derivatives of \( t \) to form a column and integrals of \( e^t \) in another, apply the down-diagonal multiplication scheme and alternate signs for the result.

Key Concepts

Integration by Parts

Integration by Parts is a valuable technique for finding integrals of functions multiplied together, known as products. When you come across a product of two functions in an integral, this method can simplify the problem significantly. Let’s break down the steps involved:

• The formula for integration by parts is: \( \int u \, dv = uv - \int v \, du \). This equation transforms a complex integral into a simpler one.
• First, you need to identify your 'u' and 'dv'. Choose 'u' such that it simplifies when differentiated, and 'dv' should be easy to integrate. For example, in our integral \( \int t e^t \, dt \), let \( u = t \), making \( dv = e^t \, dt \).
• Differentiate \( u \) to find \( du \) and integrate \( dv \) to find \( v \). This gives you \( du = dt \) and \( v = e^t \).
• Now apply the formula: substitute and simplify to solve the integral.

This method requires practice but is powerful for integrals where two functions are multiplied.

Definite Integrals

A definite integral helps us find the accumulation of quantities, like area under a curve, over a specific interval. It's a great way to understand how a function behaves across a range. Here’s how it works:

• In definite integrals, the integration process includes limits, commonly called bounds, which specify the start and end points of integration.
• The result isn't an expression involving constants but a number that signifies the net area between the curve and the x-axis for boundary values.
• The calculation involves evaluating the antiderivative at the upper boundary and subtracting the value of the same antiderivative at the lower boundary.

When using integration by parts for definite integrals, you must apply the limits to both parts of the integration by parts formula to find a precise solution within the given bounds.

Indefinite Integrals

Indefinite integrals are all about finding a family of functions whose derivative is the given function. They are essential in calculus because they deal with general forms.

• Unlike definite integrals, indefinite integrals do not have boundaries and thus represent an entire family of antiderivatives plus a constant of integration, \( C \).
• The process gives you the original function when differentiated, essentially reversing differentiation.
• Whenever you evaluate an indefinite integral, you include \( C \) to account for all possible antiderivatives. This constant illustrates that there are infinite solutions differing by a constant value.

Indefinite integrals offer a glimpse into how functions change, revealing the broader picture rather than just solid numbers, making them a fundamental part of calculus. By recognizing functions that can be easily integrated, like polynomial functions, you can swiftly find their indefinite integral.