Problem 22 - Applied Calculus

Question

Use integration by parts twice to evaluate the integral. $$\int t^{2} e^{5 t} d t$$

Step-by-Step Solution

Answer

The integral is \( e^{5t} \left( \frac{1}{5} t^2 - \frac{2}{25} t + \frac{2}{125} \right) + C \).

Step 1: Choose functions for integration by parts

For integration by parts, we choose two functions based on the integral \( \int t^2 e^{5t} dt \). Let \( u = t^2 \) and \( dv = e^{5t} dt \). Then, we need to find \( du \) and \( v \) afterwards.

Step 2: Differentiate and Integrate

Differentiate \( u = t^2 \) to find \( du = 2t \, dt \). Integrate \( dv = e^{5t} dt \) to find \( v = \frac{1}{5} e^{5t} \).

Step 3: Apply Integration by Parts Formula

Use the formula \( \int u \, dv = uv - \int v \, du \). This gives us \( \int t^2 e^{5t} dt = t^2 \cdot \frac{1}{5} e^{5t} - \int \frac{1}{5} e^{5t} \cdot 2t \, dt \). Simplify to get \( \int t^2 e^{5t} dt = \frac{1}{5} t^2 e^{5t} - \frac{2}{5} \int t e^{5t} dt \).

Step 4: Apply Integration by Parts Again

Now focus on \( \int t e^{5t} dt \). Choose new functions for integration by parts: \( u = t \), \( dv = e^{5t} dt \). Then \( du = dt \) and \( v = \frac{1}{5} e^{5t} \). Apply the formula again: \( \int t e^{5t} dt = t \cdot \frac{1}{5} e^{5t} - \int \frac{1}{5} e^{5t} \, dt \). Simplify to get \( \int t e^{5t} dt = \frac{1}{5} t e^{5t} - \frac{1}{25} e^{5t} \).

Step 5: Substitute Back Into the Equation

Substitute the expression for \( \int t e^{5t} dt \) back into the equation from step 3. We have:\[ \int t^2 e^{5t} dt = \frac{1}{5} t^2 e^{5t} - \frac{2}{5} \left( \frac{1}{5} t e^{5t} - \frac{1}{25} e^{5t} \right) \].

Step 6: Simplify the Expression

Distribute and combine like terms: \[ \int t^2 e^{5t} dt = \frac{1}{5} t^2 e^{5t} - \frac{2}{25} t e^{5t} + \frac{2}{125} e^{5t} \]. Factor out \( e^{5t} \) to clarify the pattern:\[ \int t^2 e^{5t} dt = e^{5t} \left( \frac{1}{5} t^2 - \frac{2}{25} t + \frac{2}{125} \right) + C \].

Key Concepts

Integral Calculus

Integral calculus is a fundamental branch of mathematics that focuses on the concept of integration. Integration allows us to find the area under a curve or to compute quantities like mass, area, and volume. In this exercise, we use a technique called integration by parts, which is particularly useful for integrating the products of functions.

• The integration by parts formula is derived from the product rule for differentiation: \( \int u \, dv = uv - \int v \, du \).
• This formula helps break down a complex integral into simpler parts by strategically selecting which part of the integrand to differentiate and which part to integrate.

To successfully use integration by parts, choose \( u \) and \( dv \) such that the resulting integral \( \int v \, du \) is easier to solve. This process might require applying the technique multiple times, as demonstrated in the step-by-step solution to solve \( \int t^2 e^{5t} \, dt \). Through these steps, we simplify the integral to find its solution.

Exponential Functions

Exponential functions are one of the most significant categories of mathematical functions and appear frequently in calculus problems. In the context of integration, the exponential function \( e^{5t} \) is one of the terms integrated in the given exercise.

• The function \( e^x \) has the unique property that it is its own derivative: \( \frac{d}{dx} e^x = e^x \).
• This property makes exponential functions straightforward to integrate and differentiate.

In our problem, we deal with \( e^{5t} \) and utilize the scaling rule for integrals. When integrating \( e^{at} \), where \( a \) is a constant, the primitive is \( \frac{1}{a} e^{at} + C \). This characteristic was used twice in the exercise when selecting \( dv \) and finding \( v \) after integration. Understanding this rule simplifies the process of integration by parts, as seen when \( v = \frac{1}{5} e^{5t} \) was chosen.

Polynomial Functions

Polynomial functions are indispensable in calculus due to their straightforward structure and properties. These functions are expressed as sums of terms, each consisting of a variable raised to a non-negative integer power multiplied by a coefficient, such as \( t^2 \) in our exercise.

• When integrating polynomials, each term of the polynomial is integrated separately by applying basic rules of integration.
• For example, for \( t^n \), the integral is \( \frac{t^{n+1}}{n+1} + C \) when \( n eq -1 \).

In the context of integration by parts, polynomial functions are often chosen as \( u \) because differentiating them reduces their degree, thereby simplifying the problem as evidenced in the given example where \( u = t^2 \). By reducing \( t^2 \) to \( 2t \), the complexity of the integral is gradually decreased, allowing for a manageable solution through successive application of integration by parts.