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Magazine Chapter 5: Problem 29
Evaluate the integrals using Part 1 of the Fundamental Theorem of Calculus. $$ \int_{1}^{4}\left(\frac{1}{\sqrt{t}}-3 \sqrt{t}\right) d t $$
Short Answer
Expert verified
The integral evaluates to -12.
Step by step solution
01
Set Up the Integral
We need to evaluate the definite integral \( \int_{1}^{4} \left( \frac{1}{\sqrt{t}} - 3 \sqrt{t} \right) dt \). This means we will find the antiderivative of the integrand function and then apply the limits of integration from 1 to 4.
02
Find the Antiderivatives
The integrand is \( \frac{1}{\sqrt{t}} - 3 \sqrt{t} \). This can be rewritten as \( t^{-1/2} - 3t^{1/2} \). The antiderivative of \( t^{-1/2} \) is \( 2t^{1/2} \) and the antiderivative of \( -3t^{1/2} \) is \( -2t^{3/2} \). Therefore, the antiderivative of the entire integrand is \( 2t^{1/2} - 2t^{3/2} \).
03
Apply the Limits of Integration
Using Part 1 of the Fundamental Theorem of Calculus, we evaluate the antiderivative at the upper limit and subtract the evaluation at the lower limit. \[ \left[ 2t^{1/2} - 2t^{3/2} \right]_{1}^{4} = \left( 2(4)^{1/2} - 2(4)^{3/2} \right) - \left( 2(1)^{1/2} - 2(1)^{3/2} \right) \].
04
Evaluate at the Upper Limit
Calculate the value of the antiderivative at \( t = 4 \): \( 2(4^{1/2}) - 2(4^{3/2}) = 2 \times 2 - 2 \times 8 = 4 - 16 = -12 \).
05
Evaluate at the Lower Limit
Calculate the value of the antiderivative at \( t = 1 \): \( 2(1^{1/2}) - 2(1^{3/2}) = 2 \times 1 - 2 \times 1 = 2 - 2 = 0 \).
06
Subtract Lower Limit from Upper Limit
Subtract the result of the lower limit evaluation from the upper limit evaluation: \( -12 - 0 = -12 \). Thus, the value of the integral is \(-12 \).
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Key Concepts
These are the key concepts you need to understand to accurately answer the question.
Definite Integrals
In calculus, definite integrals enable us to compute the area under a curve over a specific interval. When evaluating a definite integral like \( \int_{1}^{4} \left( \frac{1}{\sqrt{t}} - 3 \sqrt{t} \right) dt \), we use the interval from 1 to 4 as our bounds. This calculates the exact net area between the curve of the function \( \frac{1}{\sqrt{t}} - 3 \sqrt{t} \) and the \( t \)-axis over this range.
- **Upper and Lower Limits:** The numbers 1 and 4 are the lower and upper bounds of our integral, respectively.
- **Calculation:** To find the definite integral, we first find an antiderivative and then apply these limits using the Fundamental Theorem of Calculus.
- **Result:** The result of a definite integral is a specific number that represents the signed area, in this case, \(-12\), indicating not just magnitude but direction (negative suggests below the \( t \)-axis).
Antiderivatives
Antiderivatives play a crucial role in solving integrals. When integrating the function \( \frac{1}{\sqrt{t}} - 3 \sqrt{t} \), we need to find the antiderivative of each term separately.
- **Function Transformation:** Rewriting helps: \( \frac{1}{\sqrt{t}} \) becomes \( t^{-1/2} \), and \( 3 \sqrt{t} \) becomes \( 3t^{1/2} \).
- **Finding Antiderivatives:** From basic integral rules, the antiderivative of \( t^{-1/2} \) is \( 2t^{1/2} \) and that of \( -3t^{1/2} \) is \( -2t^{3/2} \).
- **Combining Results:** The antiderivative of the entire function becomes \( 2t^{1/2} - 2t^{3/2} \).
Integration Techniques
Integration techniques simplify the evaluation of complex integrals. In this example, recognizing simple power functions permits straightforward integration. Here are some basic techniques applied:
- **Power Rule for Integrals:** When dealing with expressions like \( t^n \), the antiderivative is \( \frac{t^{n+1}}{n+1} + C \), provided \( n eq -1 \).
- **Handling Roots and Powers:** By rewriting roots as powers, such as \( \sqrt{t} = t^{1/2} \), it becomes easier to apply the power rule.
- **Subtraction Between Functions:** The integrals of \( \frac{1}{\sqrt{t}} \) and \( -3 \sqrt{t} \) are calculated separately and then combined. Always check if you can simplify before integrating.
