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Chapter 5: Problem 51

Find the area under the graph over the indicated interval. $$ y=5-x^{2} ; \quad[-1,2] $$

Short Answer

Expert verified
The area under the graph is 12.

Step by step solution

01

- Set up the definite integral

To find the area under the graph of the function over the given interval, set up the definite integral from \(-1\) to \(2\) of the function \(5-x^2\). The integral to evaluate is: \[\int_{-1}^{2} (5 - x^2) \, dx\]
02

- Integrate the function

Compute the antiderivative of the integrand \(5 - x^2\). This gives: \[\int (5 - x^2) \, dx = 5x - \frac{x^3}{3}\]
03

- Evaluate the definite integral

Evaluate the antiderivative from \(-1\) to \(2\): \[\left[5x - \frac{x^3}{3}\right]_{-1}^{2}\] Substitute the upper limit (\(2\)) and lower limit (\(-1\)) into the antiderivative: \[\left(5(2) - \frac{2^3}{3}\right) - \left(5(-1) - \frac{(-1)^3}{3}\right)\] Simplify the expression: \[(10 - \frac{8}{3}) - (-5 + \frac{1}{3})\] Combine the terms: \[\left(10 - \frac{8}{3}\right) + \left(5 - \frac{1}{3}\right)\] Convert to a common denominator and sum: \[\frac{30}{3} - \frac{8}{3} + \frac{15}{3} - \frac{1}{3} = \frac{36}{3} = 12\]

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Key Concepts

These are the key concepts you need to understand to accurately answer the question.

Area Under the Curve
The area under the curve represents the region between the graph of a function and the x-axis. In this problem, our goal is to find the area under the curve of the function \(y = 5 - x^2\) between the points \(x = -1\) and \(x = 2\). This area can be found using definite integrals. The concept is key in calculus because it helps in measuring the total quantity under a curve, which can represent various physical quantities like distance, probability, or volume.
Antiderivative
To find the area under a curve, we use the antiderivative of the function, also known as the integral. The antiderivative of a function \(f(x)\) is another function \(F(x)\) such that the derivative \(F'(x)\) equals \(f(x)\). For our function \(5 - x^2\), the antiderivative is \(5x - \frac{x^3}{3}\). This new function, evaluated at the bounds of the integral, helps us calculate the total area under the curve.
Evaluating Integrals
After finding the antiderivative, the next step is evaluating the definite integral over the given bounds. This means we calculate the antiderivative at the upper limit and subtract the value of the antiderivative at the lower limit. With our function, we evaluate from \(-1\) to \(2\) as follows:
  • Calculate \(5(2) - \frac{2^3}{3}\)
  • Then calculate \(5(-1) - \frac{(-1)^3}{3}\)
  • Finally, subtract the second result from the first.
The expression we get helps us find the exact area.
Integral Bounds
Integral bounds are the values at which we start and stop the integration process. In this exercise, the bounds are \(-1\) and \(2\). These bounds help us limit our area calculation to a specific interval along the x-axis. Setting up the integral with these bounds, we specify exactly where to start and end on the curve \(y = 5 - x^2\). The integration process sums up the infinitesimally small areas from the lower bound to the upper bound, giving us the total area under the curve for this interval.

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