Problem 5 - Calculus and its applications

Question

Find the area under the graph of \(f\) over [-6,4] . $$ f(x)=\left\\{\begin{array}{lll} -x^{2}-6 x+7, & \text { for } & x<1 \\ \frac{3}{2} x-1, & \text { for } & x \geq 1 \end{array}\right. $$

Step-by-Step Solution

Answer

Area under the graph over [-6,4] is 112.25.

Step 1: Identify the Regions

The function is defined piecewise, splitting at \( x=1 \). Therefore, we need to consider two separate regions: Region A for \([-6, 1)\) and Region B for \([1, 4]\).

Step 2: Set Up the Integral for Region A

For \( x < 1 \), use the expression \( f(x) = -x^2 - 6x + 7 \). The area under the curve from \(-6\) to \(1\) is given by \( \int_{-6}^{1} (-x^2 - 6x + 7) \, dx \).

Step 3: Integrate the Function for Region A

Integrate \( -x^2 - 6x + 7 \): \[\int (-x^2 - 6x + 7) \, dx = \left[-\frac{x^3}{3} - 3x^2 + 7x\right] + C\].Evaluate it from \(-6\) to \(1\):\[-\left( \frac{1^3}{3} + 3 \times 1^2 - 7 \times 1 \right) - \left( -\frac{-6^3}{3} - 3 \times (-6)^2 + 7 \times (-6) \right) \].

Step 4: Calculate the Area for Region A

Evaluate \[= -\left( \frac{1}{3} + 3 - 7 \right) - \left( -\frac{-216}{3} - 3 \times 36 + 7 \times -6 \right)\] \[= -\left( -\frac{10}{3} \right) - \left( 72 - 108 - 42 \right)\] \[= \frac{10}{3} + 78 = \frac{234}{3} + 78 = \frac{78}{3} = 104\].

Step 5: Set Up the Integral for Region B

For \( x \geq 1 \), use the expression \( f(x) = \frac{3}{2} x - 1 \). The area from \(1\) to \(4\) is given by \( \int_{1}^{4} \left( \frac{3}{2} x - 1 \right) \, dx \).

Step 6: Integrate the Function for Region B

Integrate:\[\int \left( \frac{3}{2} x - 1 \right) \, dx = \left[ \frac{3}{4}x^2 - x \right] + C\].Evaluate it from \(1\) to \(4\):\[= \left( \frac{3}{4} \times 4^2 - 4 \right) - \left( \frac{3}{4} \times 1^2 - 1 \right)\].

Step 7: Calculate the Area for Region B

Evaluate:\[= \left( \frac{3}{4} \times 16 - 4 \right) - \left( \frac{3}{4} - 1 \right)\] \[= (12 - 4) - \left( \frac{3}{4} - 1 \right) = 8 - \left( -\frac{1}{4} \right)\] \[= 8 + \frac{1}{4} = \frac{32}{4} + \frac{1}{4} = \frac{33}{4} = 8.25\].

Step 8: Combine the Areas to Find Total Area

Add the areas of both regions to find the total area under the graph:\[104 + 8.25 = 112.25\].

Key Concepts

Definite Integral

The concept of a 'Definite Integral' in calculus represents the accumulation of quantities, which can be thought of as the net area under a curve within a specific interval. In our exercise, this involves calculating the area under the function \( f(x) \) from \( x = -6 \) to \( x = 4 \). The definite integral is expressed as \( \int_{a}^{b} f(x) \, dx \), where \( a \) and \( b \) are the lower and upper limits of the interval. To solve this, you find the antiderivative of the function, evaluate it at the boundaries \( a \) and \( b \), and compute the difference. This provides an exact value of the area, or if the curve lies below the x-axis, it gives the negative of the area.

Piecewise Functions

A 'Piecewise Function' is a function defined by multiple sub-functions, each applying to a certain interval of the main function's domain. In the given problem, the function \( f(x) \) is defined differently for values below or equal to 1 and above 1. This requires us to handle each segment separately, as different expressions govern different parts of the x-axis. The role of piecewise functions in integration involves correctly setting up separate integrals for each segment, respecting the function's structure and limits of integration.

Area under a Curve

Finding the area under a curve is a central application of integral calculus. It involves computing the definite integral of a function over a specific interval. The area under the graph of our piecewise function \( f(x) \), from \(-6\) to \(4\), sums the areas of two regions defined by the different pieces of \( f(x) \). Since both parts of the function contribute unique expressions, the problem requires evaluating separate integrals for each region. Summing these integral values yields the total area which represents the entire region enclosed by the curve and the x-axis.

Integration Techniques

Various 'Integration Techniques' are employed to find the integral of a function, particularly when dealing with polynomial and simple linear functions. A common technique involves integrating each term separately using basic integration rules: power rule for polynomials \( x^n \rightarrow \frac{x^{n+1}}{n+1} \) and constant multiple rule. After integrating, evaluate the antiderivative at the endpoints of the interval. It’s crucial to apply these techniques separately for each sub-function within a piecewise function, making sure to correctly handle limits where the function changes. This guarantees an accurate calculation of the area under the curve. Integrating correctly ensures that all contributions of the original function across the specified intervals are accounted for.