Problem 17 - Calculus: Early Transcendental Functions

Question

Write the given (total) area as an integral or sum of integrals. The area between \(y=x^{3}-3 x^{2}+2 x\) and the \(x\) -axis for \(0 \leq x \leq 2\).

Step-by-Step Solution

Answer

The area between the curve and the x-axis from \(x=0\) to \(x=2\) is \(\frac{7}{4}\).

Step 1: Setting up the Integral

You can express the area \(A\) between the curve and the x-axis from \(x=0\) to \(x=2\) as the definite integral \[A=\int_{0}^{2} |x^{3}-3 x^{2}+2 x| dx\] It's important to use absolute value because we want to count area below the x-axis as positive.

Step 2: Splitting into Multiple Integrals

The integrand changes sign at \(x=0, 1, 2\). Therefore split the integral at these points: \[A=\int_{0}^{1} (x^{3}-3 x^{2}+2 x) dx - \int_{1}^{2} (x^{3}-3 x^{2}+2 x) dx\]

Step 3: Evaluating the Integrals

Evaluate these integrals separately. The integral of \(x^{3}-3 x^{2}+2 x\) is \(\frac{1}{4}x^4-x^3+x^2)\).Substitute the limits of the first integral:\[\int_{0}^{1} (x^{3}-3 x^{2}+2 x) dx = [\frac{1}{4}*1^4 - 1*1^3 + 1*1^2] - [\frac{1}{4}*0^4 - 0^3 + 0^2] = -\frac{3}{4}\]Substitute the limits of the second integral:\[-\int_{1}^{2} (x^{3}-3 x^{2}+2 x) dx = -[\frac{1}{4}*2^4-2^3+2*2^2 - (\frac{1}{4}*1^4-1^3+1^2)] = -1 \]

Step 4: Sum the results

Add the two results from the integral evaluations together to find the total area: \[ A = -\frac{3}{4} - 1 = -\frac{7}{4}\]But since we are interested in the absolute value (as areas are always positive), the area is \[ A = |\frac{-7}{4}| = \frac{7}{4} \]

Key Concepts

Absolute Value in Integrals

When computing the area between a curve and the x-axis using definite integrals, it's essential to consider that areas below the x-axis are normally accounted for as negative. However, when calculating the total area enclosed by the curve and the axis, we are often interested in the absolute area, without regard to whether the curve lies above or below the x-axis. To address this, we use the absolute value function within the integral.

Splitting Integrals at Points of Sign Change

In order to correctly calculate the area, particularly when dealing with a function that changes sign over the interval of integration, we need to partition the integral into segments where the function does not change sign. This process is known as splitting the integral at points of sign change. By doing so, we can then apply the absolute value correctly to each segment, ensuring that we are summing positive quantities to get the total area.

Integral Evaluation with Limits

Evaluation of definite integrals with limits involves calculating the antiderivative of the function, then applying the Fundamental Theorem of Calculus. This means substituting the upper limit into the antiderivative, subtracting the result from substituting the lower limit, and finding the difference between them. The process is repeated for each segment if the integral has been split due to sign change in the function. The individual results are then summed to obtain the total area covered by the curve.

Area Between Curve and X-axis

The area between a curve defined by a function and the x-axis can either be above the axis, below the axis, or both. When the graph is above the x-axis the area is positive, and when it is below, the area calculated from the integral appears as negative. However, to find the magnitude of the total area—regardless of position relative to the x-axis—we consider the absolute value of the function within the integral over the given interval. This ensures all segments of the area are aggregated as positive values.