Problem 55 - Calculus Early Transcendentals

Question

Find the area of the region bounded by the graph of \(f\) and the \(x\) -axis on the given interval. $$f(x)=x^{2}-25 \text { on }[2,4]$$

Step-by-Step Solution

Answer

The area of the region bounded by the graph and the x-axis on the interval \([2, 4]\) is \(\frac{86}{3}\).

Step 1: Determine if \(f(x)\) is above or below the x-axis

For this, we will evaluate \(f(x)\) at the given endpoints of the interval, \(x=2\) and \(x=4\). $$f(2)=(2)^2-25=4-25=-21$$ $$f(4)=(4)^2-25=16-25=-9$$ Since both function values are negative, this tells us that the graph of \(f(x)\) lies below the x-axis on the interval \([2,4]\).

Step 2: Set up the area formula using integrals

To find the area, we will use the formula: $$A=\int_{a}^{b}{|f(x)|} dx$$ In our case, \(f(x)\) is negative on the interval \([2,4]\). Therefore, the absolute value of \(f(x)\) becomes: $$|f(x)|=|x^2-25|=-(x^2-25)=25-x^2$$ Now we can set up the integral to find the area.

Step 3: Evaluate the integral

Now, we can evaluate the integral: $$A=\int_{2}^{4}(25-x^2)dx$$ To solve the integral, we will use the power rule for integration: $$\int (25-x^2)dx = 25x - \frac{1}{3}x^3 + C$$ Now, evaluate the antiderivative at the limits 2 and 4, and subtract to find the area: $$A=[25(4)-\frac{1}{3}(4)^3] - [25(2)-\frac{1}{3}(2)^3] = [100-\frac{64}{3}] - [50 - \frac{8}{3}] = 50 - \frac{56}{3} = \frac{86}{3}$$ So the area of the region bounded by the graph of \(f(x)=x^2-25\) and the x-axis on the interval \([2, 4]\) is \(\frac{86}{3}\).

Key Concepts

Definite Integrals

When it comes to finding the area under a curve on a specific interval, definite integrals are the tool of choice. A definite integral is represented by the integral sign with upper and lower limits, which indicate the bounds of the interval. In the context of our exercise, the integral is used to calculate the area of the region bounded by the function f(x) = x^2 - 25 and the x-axis between x = 2 and x = 4.

The process involves taking the integral of the absolute value of the function to ensure that we are considering the correct 'above' or 'below' the x-axis distinction. The calculation of a definite integral results in a number that represents the net area: the area above the x-axis minus the area below the x-axis. In our case, since the curve lies entirely below the x-axis on the interval, we need to integrate the negative of the function to find the positive area value.

Integration by Power Rule

The power rule for integration is straightforward and a fundamental technique in calculus. It provides a method to integrate polynomials term by term. The rule states that for any real number n not equal to -1, the integral of x^n with respect to x is (x^(n+1))/(n+1), plus the constant of integration, C.

In practice, if we use this rule on our function f(x), each term in the polynomial (25 - x^2) is integrated separately. Hence, the integral of 25 with respect to x becomes 25x, and the integral of -x^2 becomes -x^3/3. A constant of integration is usually included, but it cancels out when evaluating a definite integral. By applying this rule, we find the function that gives us the accumulated area under the curve of our original function.

Absolute Value of Function

When determining the area between a function and the x-axis, the absolute value plays a pivotal role, especially if the function crosses the x-axis. The absolute value of a function, denoted as |f(x)|, ensures that all values are taken as positive, which is essential for calculating areas, as they cannot be negative.

In the given problem, the function f(x) = x^2 - 25 is negative over the interval [2,4]. To address this when finding the area, we take the absolute value of the function, effectively reflecting the part of the graph below the x-axis up above the x-axis for the purpose of area calculation. For our function, this translates to using -(x^2 - 25) instead of x^2 - 25 in the integration process. This ensures that the computed area is positive, as it should be in geometric contexts.