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Magazine Chapter 11: Problem 33
Evaluate the definite integral. $$\int_{1}^{3}\left(t^{2}-t\right)^{2} d t$$
Short Answer
Expert verified
The value of the definite integral \(\int_{1}^{3}\left(t^{2}-t\right)^{2} d t\) is \(-\frac{223}{30}\).
Step by step solution
01
Expand the Integrand
First, expand the expression \((t^2 - t)^2\) using the formula (a - b)^2 = (a^2 - 2ab + b^2):
$$\int_{1}^{3} \left(t^{2}-t\right)^{2} dt = \int_{1}^{3} \left(t^4 - 2t^3 + t^2\right) dt$$
02
Integrate the Expanded Function
Now, integrate each term of the expanded function using the Power Rule of integration, which states that \(\int x^n dx = x^{n+1} / (n+1) + C\):
$$\int_{1}^{3} \left(t^4 - 2t^3 + t^2\right) dt = \left[\frac{t^5}{5} - \frac{2t^4}{4} + \frac{t^3}{3}\right]_{1}^{3}$$
03
Evaluate the Integral over the Interval
Finally, to evaluate the definite integral, substitute the upper limit (3) for t and subtract the result of substituting the lower limit (1) for t:
$$\left[\frac{t^5}{5} - \frac{2t^4}{4} + \frac{t^3}{3}\right]_{1}^{3} = \left(\frac{3^5}{5} - \frac{2(3^4)}{4} + \frac{3^3}{3}\right) - \left(\frac{1^5}{5} - \frac{2(1^4)}{4} + \frac{1^3}{3}\right)$$
Calculate the values:
$$= \left(\frac{243}{5} - \frac{162}{2} + 9\right) - \left(\frac{1}{5} - \frac{2}{4} + \frac{1}{3}\right)$$
$$= \left(\frac{243}{5} + \frac{9}{3} - \frac{162}{2}\right) - \left(\frac{1}{5} + \frac{1}{3} - \frac{1}{2}\right)$$
$$= \frac{486 + 90 - 810}{10} - \frac{6 - 10 + 15}{30}$$
$$= \frac{-234}{10} + \frac{11}{30}$$
$$= \frac{-468}{60} + \frac{22}{60}$$
$$= \frac{-446}{60} = -\frac{223}{30}$$
So, the value of the definite integral is:
$$\int_{1}^{3}\left(t^{2}-t\right)^{2} d t = -\frac{223}{30}$$
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Key Concepts
These are the key concepts you need to understand to accurately answer the question.
Integration by Expansion
When faced with a complex expression within an integral, one effective strategy is integration by expansion. This involves simplifying the integrand by expanding it into a series of simpler terms that are easier to integrate. In the context of our exercise, the integrand \( (t^2 - t)^2 \) was expanded using the algebraic identity \( (a - b)^2 = a^2 - 2ab + b^2 \), resulting in a polynomial \( t^4 - 2t^3 + t^2 \) that is much simpler to integrate term by term.
Expansion becomes particularly useful when dealing with polynomials raised to a power, which is common in calculus problems. By expanding such terms, students can make use of simpler integration rules, like the power rule, instead of attempting to directly integrate the original, more complicated function.
Expansion becomes particularly useful when dealing with polynomials raised to a power, which is common in calculus problems. By expanding such terms, students can make use of simpler integration rules, like the power rule, instead of attempting to directly integrate the original, more complicated function.
Power Rule of Integration
A fundamental tool in evaluating integrals of polynomial expressions is the power rule of integration. The power rule states that for any real constant n that is not equal to -1, the integral of \( x^n \) with respect to \( x \) is \( \frac{x^{n+1}}{n+1} + C \) where \( C \) represents the constant of integration. In definite integrals, the constant C is omitted since it cancels out when the limits are applied.
For example, in the given exercise, the expanded polynomial \( t^4 - 2t^3 + t^2 \) is integrated term by term using the power rule. This yields an antiderivative \( \frac{t^5}{5} - \frac{2t^4}{4} + \frac{t^3}{3} \) which can then be evaluated from the lower to the upper limit of the integral.
For example, in the given exercise, the expanded polynomial \( t^4 - 2t^3 + t^2 \) is integrated term by term using the power rule. This yields an antiderivative \( \frac{t^5}{5} - \frac{2t^4}{4} + \frac{t^3}{3} \) which can then be evaluated from the lower to the upper limit of the integral.
Evaluate Definite Integral
To evaluate a definite integral, one must find the antiderivative of the integrand and then apply the Fundamental Theorem of Calculus, which tells us to subtract the value of the antiderivative at the lower limit from its value at the upper limit. This process turns a problem of finding the area under a curve into a straightforward arithmetic calculation.
In our exercise, after obtaining the antiderivative, we substitute the upper limit (3) and lower limit (1), taking the difference of these two values to find the integral's value over the interval [1, 3]. The result is a single number that represents the exact area under the curve of the function \( (t^2 - t)^2 \) between these two points.
In our exercise, after obtaining the antiderivative, we substitute the upper limit (3) and lower limit (1), taking the difference of these two values to find the integral's value over the interval [1, 3]. The result is a single number that represents the exact area under the curve of the function \( (t^2 - t)^2 \) between these two points.
Polynomial Integration
The integration of polynomial functions like \( t^n \) is streamlined by using the power rule. Polynomial integration refers to the process of integrating polynomial expressions, which are sums of variables raised to whole number powers. Each term of a polynomial is integrated separately, resulting in an antiderivative that is itself a polynomial.
In our textbook exercise, we deal with a polynomial function after expanding the original expression. The power rule is applied to integrate each term of the polynomial separately, demonstrating a key technique in calculus for dealing with polynomial integrands. This method is widely applicable because many functions can be approximated by polynomials, and thus polynomial integration is a crucial concept in understanding broader applications of integration.
In our textbook exercise, we deal with a polynomial function after expanding the original expression. The power rule is applied to integrate each term of the polynomial separately, demonstrating a key technique in calculus for dealing with polynomial integrands. This method is widely applicable because many functions can be approximated by polynomials, and thus polynomial integration is a crucial concept in understanding broader applications of integration.
