Chapter 22: Problem 150
If \(f(y)=e^{y}, g(y)=y ; y>0\) and \(F(t)=\int_{0}^{t} f(t-y) g(y) d y\), then (a) \(F(t)=t e^{-t}\) (b) \(F(t)=1-t e^{-t}(1+t)\) (c) \(F(t)=e^{t}-(1+t)\) (d) \(F(t)=t e^{t} .\)
Short Answer
Expert verified
Option (b) is correct: \( F(t) = 1 - t e^{-t} (1 + t) \).
Step by step solution
01
Rewrite the Function Substitutions
Let's first understand the expression for \( F(t) \). We are given that:\[ F(t) = \int_{0}^{t} f(t-y) g(y) \, dy \]Where \( f(y) = e^{y} \) and \( g(y) = y \). So, \( f(t-y) = e^{t-y} \) and \( g(y) = y \) becomes:\[ F(t) = \int_{0}^{t} e^{t-y} y \, dy \]
02
Simplify the Expression
We can simplify the integral expression further:\[ F(t) = \int_{0}^{t} e^{t-y} y \, dy = e^{t} \int_{0}^{t} e^{-y} y \, dy \]The constant \( e^{t} \) can be factored out of the integral.
03
Integrate by Parts
To find \( \int e^{-y} y \, dy \), use integration by parts:Let \( u = y \) and \( dv = e^{-y} \, dy \).Then \( du = dy \) and \( v = -e^{-y} \).The integration by parts formula is \( \int u \, dv = uv - \int v \, du \).Substitute to get:\[ \int y e^{-y} \, dy = -y e^{-y} - \int (-e^{-y}) \, dy \]Simplifying further, you get:\[ \int y e^{-y} \, dy = -y e^{-y} + e^{-y} \]
04
Evaluate Definite Integral
Evaluate the definite integral from 0 to \( t \):\[ \left[ -y e^{-y} + e^{-y} \right]_0^t = \left(-t e^{-t} + e^{-t}\right) - \left(0 + 1\right) \]This simplifies to:\[ -t e^{-t} + e^{-t} - 1 \]
05
Substitute Back to Expression for F(t)
Recall from Step 2 that:\[ F(t) = e^{t} \left( -t e^{-t} + e^{-t} - 1 \right) \]Simplify this expression:\[ F(t) = (1 - t e^{-t}) \]Checking all given options, the expression can be simplified to match one of the choices.
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Key Concepts
These are the key concepts you need to understand to accurately answer the question.
Definite Integral
The concept of a definite integral is a crucial building block in calculus. A definite integral represents the area under the curve of a function, between two points on the x-axis. In mathematical notation, the definite integral of a function \( f(x) \) from \( a \) to \( b \) is written as \( \int_a^b f(x) \, dx \). It calculates the accumulated quantity, which can be thought of as the total change over a certain interval.
When working with definite integrals, it's important to:
When working with definite integrals, it's important to:
- Identify the upper and lower limits of integration, which are the boundaries over which you are accumulating the function's values.
- Understand that the definite integral provides a numerical value, not another function. This distinguishes it from an indefinite integral, which results in a family of functions.
- Be mindful of the properties of integrals such as the additive property: \( \int_a^b f(x) \, dx = \int_a^c f(x) \, dx + \int_c^b f(x) \, dx \).
Exponential Functions
Exponential functions are a special type of function where the variable is in the exponent. A basic form is \( f(x) = a^x \), where \( a \) is a constant and \( x \) is an exponent. The constant \( a \) is typically a positive real number, and if \( a = e \), the function becomes \( f(x) = e^x \), which is prevalent in many areas of science and mathematics.
Key properties of exponential functions include:
Key properties of exponential functions include:
- Rapid growth or decay, depending on whether the base \( a \) is greater or less than 1.
- The function \( e^x \) is its own derivative, which is a unique and powerful property, making it invaluable in calculus.
- Incorporating these functions into integrals often leads to elegant solutions, as they interact beautifully with different calculus rules, especially integration by parts.
Function Integration
Function integration is a fundamental process in calculus used to find the integral, or the area under the curve, of a function. It can either be a definite or an indefinite calculation. While a definite integral has specific limits, an indefinite integration results in a general form or family of antiderivatives of the function.
Function integration involves:
Function integration, especially when involving two or more functions, requires careful handling of each component, as seen in our exercise where we integrated \( e^{-y}y \) using parts to arrive at the needed solution.
Function integration involves:
- Finding the antiderivative: Reverse the process of differentiation to uncover the original function.
- Using techniques like substitution and integration by parts to break down complex integrals into manageable steps.
- Applying limits in definite integrals to get a specific numerical result.
Function integration, especially when involving two or more functions, requires careful handling of each component, as seen in our exercise where we integrated \( e^{-y}y \) using parts to arrive at the needed solution.