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Magazine Chapter 10: Problem 55
Evaluate the integrals . $$ \int_{0}^{1}\left(x^{2}-4 y\right) d x, \text { where } x=t+1, y=t^{3}+4 $$
Short Answer
Expert verified
The integral evaluates to \( \frac{44}{3} \).
Step by step solution
01
Variable Substitution
We are given the integrals with the substitutions that are needed: \( x = t + 1 \) and \( y = t^3 + 4 \). Let's substitute these expressions into the integral.
02
Substitute x and y
Substitute \( x \) and \( y \) into the integral: \[ and substitute: \ x^2 = (t+1)^2 = t^2 + 2t + 1 \ y = t^3 + 4.\\\int_0^1 \left(x^2 - 4y\right) dx = \int_0^1 ((t^2 + 2t + 1) - 4(t^3 + 4)) dt\\]
03
Simplify the Expression
Simplify the expression inside the integral: \[(t^2 + 2t + 1) - 4(t^3 + 4) = t^2 + 2t + 1 - 4t^3 - 16\]Combine like terms: \[= -4t^3 + t^2 + 2t - 15\]
04
Adjust the Integral Bounds
Determine the bounds for \( t \). When \( x = 0 \) (the lower bound on \( x \)), \( t+1 = 0 \) gives \( t = -1 \). When \( x = 1 \) (the upper bound on \( x \)), \( t+1 = 1 \) gives \( t = 0 \). Update the integral limits from \( -1 \) to \( 0 \).
05
Integrate the Function
Calculate the integral of \(-4t^3 + t^2 + 2t - 15\) from \( t = -1 \) to \( t = 0 \):\[\int_{-1}^{0} (-4t^3 + t^2 + 2t - 15) \, dt \]This becomes:\[\left[ -t^4 + \frac{t^3}{3} + t^2 - 15t \right]_{-1}^{0}\]
06
Evaluate the Integral
Calculate the value of: \[\left[ -t^4 + \frac{t^3}{3} + t^2 - 15t \right]_{-1}^{0}\] Evaluating at \( t = 0 \) yields zero. For \( t = -1 \): \[-(-1)^4 + \frac{(-1)^3}{3} + (-1)^2 - 15(-1)\]Which results in:\[-1 - \frac{1}{3} + 1 + 15 = 15 - \frac{1}{3} = \frac{44}{3}\]
07
Final Step: Conclusion of Integration
Thus, after evaluating the integral, its value is \( \frac{44}{3} \).
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Key Concepts
These are the key concepts you need to understand to accurately answer the question.
Variable Substitution
In the world of calculus integration, variable substitution is like a secret weapon. It helps simplify complex integrals by changing the variables used in the expression. In our exercise, we swap out variables from the integral with given functions of a new variable, say \( t \).
- Why Substitution?: It can convert a complicated integral into a simpler one, making it easier to evaluate.
- How it Works: Substitute the given expressions \( x = t + 1 \) and \( y = t^3 + 4 \) directly into the integral expression.
- After substitution, the integral becomes a new expression dependent on \( t \) instead of the original variables \( x \) and \( y \).
Integral Bounds
Integral bounds play a crucial role in solving definite integrals. They tell us the interval over which you need to evaluate the function. These bounds are the start and end points of integration.
- Original Bounds: The exercise starts with bounds from \( x = 0 \) to \( x = 1 \).
- Transforming Bounds: With variable substitution, the bounds also need updating. Solve \( x = t + 1 \) for \( t \) with the old bound values. When \( x = 0 \), \( t = -1 \). When \( x = 1 \), \( t = 0 \).
- Now, the bounds are from \( t = -1 \) to \( t = 0 \), perfectly aligned with the substituted expression.
Definite Integral
Definite integrals provide us with the net area under a curve bounded by the integral limits. This concept is pivotal in numerous real-world applications like calculating distance, area, and even probability.
- Expression After Substitution: Our new integral expression becomes \( \int_{-1}^{0} (-4t^3 + t^2 + 2t - 15) \, dt \).
- Solving the Integral: Integrate term by term, considering the power reduction of \( t^n \) to \( \frac{t^{n+1}}{n+1} \).
- The definite integral results in a single numerical value, achieved by evaluating the integrated function at the provided bounds.
Polynomial Integration
Polynomial integration involves integrating expressions made up of sums of powers of a variable. Each power is treated individually during integration.
- Terms to Handle: Our example includes terms like \(-4t^3, t^2, 2t, \) and \(-15 \).
- Integration Rule: For every term \( at^n \), the integral is \( \frac{a}{n+1}t^{n+1} \).
- Break down the integral into individual integrations of each term.
- Evaluate the polynomial at the integral bounds to get the final result.
