Problem 43
Question
Evaluate the following integrals using the Fundamental Theorem of Calculus. $$\int_{1}^{4}(1-x)(x-4) d x$$
Step-by-Step Solution
Verified Answer
Question: Evaluate the integral of the given expression: $$\int_{1}^{4}(1-x)(x-4) d x$$
Answer: The integral of the given expression is $$\int_{1}^{4}(1-x)(x-4) d x = -\frac{241}{6}$$
1Step 1: Expand the expression inside the integral
To solve this integral, we will first expand the expression:
$$(1-x)(x-4) = (1*x - 1*4) + (-x*x + (-x) * -4) = x - 4 - x^2 + 4x$$
The integral becomes:
$$\int_{1}^{4}(x - 4 - x^2 + 4x) d x$$
2Step 2: Integrate the expression with respect to x
Now, we will integrate the expression with respect to x:
$$\int (x - 4 - x^2 + 4x) d x = \int (5x - x^2 - 4) d x = \frac{5}{2}x^2 - \frac{1}{3}x^3 - 4x$$
3Step 3: Use the Fundamental Theorem of Calculus
Now, we'll apply the Fundamental Theorem of Calculus:
$$\left[\frac{5}{2}x^2 - \frac{1}{3}x^3 - 4x\right]_{1}^{4}$$
To calculate the value of the integral, we need to plug in the limit values and take their difference:
$$\left(\frac{5}{2}(4)^2 - \frac{1}{3}(4)^3 - 4(4)\right) - \left(\frac{5}{2}(1)^2 - \frac{1}{3}(1)^3 - 4(1)\right)$$
Now, calculate the expression:
$$\left(40 - 64 - 16\right) - \left(\frac{5}{2} - \frac{1}{3} - 4\right) = (-40) - \left(\frac{1}{6}\right)$$
So, the integral of the given expression is:
$$\int_{1}^{4}(1-x)(x-4) d x = -\frac{241}{6}$$
Key Concepts
Definite IntegralsPolynomial IntegrationEvaluation of Limits
Definite Integrals
Definite integrals are an essential part of calculus, allowing us to find the accumulated quantity, or area under a curve, between two points along the x-axis. When we calculate a definite integral, we need bounds or limits, specified at the lower and upper limits (e.g., from 1 to 4, as in our example exercise).
The notation for definite integrals is \[\int_{a}^{b} f(x) \,dx\]where \(a\) and \(b\) are the lower and upper limits, respectively, and \(f(x)\) is the function being integrated.
To solve definite integrals, we use the **Fundamental Theorem of Calculus**,which links the concept of differentiation with integration. It states that if \(F\) is an antiderivative of \(f\) on an interval \([a, b]\), then:\[\int_{a}^{b} f(x) \,dx = F(b) - F(a)\]This provides a straightforward method to evaluate integrals by finding the antiderivative of the function, evaluating it at the upper limit, and then subtracting the evaluation at the lower limit.
The notation for definite integrals is \[\int_{a}^{b} f(x) \,dx\]where \(a\) and \(b\) are the lower and upper limits, respectively, and \(f(x)\) is the function being integrated.
To solve definite integrals, we use the **Fundamental Theorem of Calculus**,which links the concept of differentiation with integration. It states that if \(F\) is an antiderivative of \(f\) on an interval \([a, b]\), then:\[\int_{a}^{b} f(x) \,dx = F(b) - F(a)\]This provides a straightforward method to evaluate integrals by finding the antiderivative of the function, evaluating it at the upper limit, and then subtracting the evaluation at the lower limit.
Polynomial Integration
Polynomial integration involves finding the antiderivative of a polynomial function. This step is essential for completing definite integrals when the function we are integrating is a polynomial, as it was in our exercise (where we had \(x - 4 - x^2 + 4x\)).
When integrating polynomial terms,
This process transforms the polynomial into a form we can easily evaluate using the Fundamental Theorem of Calculus.
When integrating polynomial terms,
- Apply the power rule for integration: If \(f(x) = x^n\), then its integral is \(\frac{x^{n+1}}{n+1} + C\), where \(C\) is the integration constant (irrelevant for definite integrals).
- Integrate term by term: For a polynomial \(ax^m + bx^n + c\), integrate each piece separately.
This process transforms the polynomial into a form we can easily evaluate using the Fundamental Theorem of Calculus.
Evaluation of Limits
Once a polynomial is integrated, the next step is to evaluate the antiderivative at the specified limits of the definite integral. This step is called **evaluation of limits**,and it determines the definite integral's value by computing the difference between the antiderivative's evaluations at the upper and lower bounds.
For our exercise,
Always remember: this subtraction accurately provides the difference between the function's values over the given interval, essential in determining the integral's result.
For our exercise,
- Calculate \(F(b)\) where \(F(x) = \frac{5}{2}x^2 - \frac{1}{3}x^3 - 4x\) at \(x = 4\).
- Calculate \(F(a)\) where \(F(x)\) at \(x = 1\).
- Subtract these two values: \(F(b) - F(a)\).
Always remember: this subtraction accurately provides the difference between the function's values over the given interval, essential in determining the integral's result.
Other exercises in this chapter
Problem 43
Use a change of variables to evaluate the following definite integrals. $$\int_{-1}^{2} x^{2} e^{x^{3}+1} d x$$
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Use symmetry to evaluate the following integrals. $$\int_{-\pi / 4}^{\pi / 4} \sec ^{2} x d x$$
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Riemann sums for larger values of \(n\) Complete the following steps for the given function \(f\) and interval. a. For the given value of \(n\), use sigma notat
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Suppose \(\int_{0}^{3} f(x) d x=2, \int_{3}^{6} f(x) d x=-5\) and \(\int_{3}^{6} g(x) d x=1 .\) Evaluate the following integrals. a. \(\int_{0}^{3} 5 f(x) d x \
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