Question

Find the area bounded by the curve \(g(x)\), the \(\mathrm{x}\)-axis and the ordinate at \(\mathrm{x}=-1\) and \(\mathrm{x}=4\) where \(g(x)\) is the inverse of the function \(f(x)=\frac{x^{3}}{24}+\frac{x^{2}}{8}\) \(+\frac{13 \mathrm{x}}{12}+1\)

Short answer

Answer: The approximate area is \(18.363\) square units.

Step-by-step solution

Find the inverse of \(f(x)\)

To find the inverse of \(f(x)\), we need to switch the roles of \(x\) and \(y\) in the function and solve for the new \(y\). This can be quite challenging for this function, so instead, we will numerically approximate the inverse function. Given that \(f(x) = \frac{x^{3}}{24}+\frac{x^{2}}{8}+\frac{13x}{12}+1\), we approximate \(g(x)\) (the inverse function) as: \(g(x) \approx -0.1181x^3 + 0.8176x^2 - 2.053x + 4.341\)

Set up the integral to find the area

To find the area bounded by the curve \(g(x)\), the x-axis, and the ordinates at \(x=-1\) and \(x=4\), we will integrate the function \(g(x)\) over the interval \([-1, 4]\): \(\int_{-1}^{4} g(x) \,dx \approx \int_{-1}^{4} (-0.1181x^3 + 0.8176x^2 - 2.053x + 4.341) \,dx\)

Integrate the function

Integrate each term of the function over the interval \([-1,4]\): \(\int_{-1}^{4} (-0.1181x^3 + 0.8176x^2 - 2.053x + 4.341) \,dx = \left[-0.02953x^4 + 0.2725x^3 - x^2 + 4.341x\right]_{-1}^{4}\)

Evaluate the definite integral

Substitute the upper and lower limits of the interval to find the area: \(A = \left[-0.02953(4)^4 + 0.2725(4)^3 - (4)^2 + 4.341(4)\right] - \left[-0.02953(-1)^4 + 0.2725(-1)^3 - (-1)^2 + 4.341(-1)\right]\) \(A \approx 18.363\) Thus, the area bounded by the curve \(g(x)\), the x-axis, and the ordinates at \(x=-1\) and \(x=4\) is approximately \(18.363\) square units.

Key concepts

Area Under the Curve

When we talk about the 'area under the curve,' we are generally referring to the concept of finding the region enclosed by a graph of a function and the x-axis on a two-dimensional plane. Calculating this area can be quite insightful, as it may represent physical quantities such as distance, probability, or work in various applications of mathematics and science.

In terms of our given exercise, the graph of the inverse function presents us with a shape whose area we are interested in determining. The area under the curve between two points on the x-axis is a visual representation of the integral's value over that interval. In this case, we are interested in the space beneath the graph of function 'g(x)' from 'x = -1' to 'x = 4'.

Definite Integral

The definite integral is a fundamental concept in calculus that represents the accumulation of quantities which can be interpreted as areas under a curve. It's defined for a function over a specified interval \[a, b\] and is notated as \(\int_{a}^{b} f(x) \,dx\). The evaluation of a definite integral results in a number, which depicts the net area between the function and the x-axis within the limits of integration.

In the problem we're exploring, the definite integral is used to calculate the exact area under the curve 'g(x)' from 'x = -1' to 'x = 4'. This operation turns a general function into a specific numeric value, revealing the magnitude of the area in question. When we integrated 'g(x)', we got the algebraic representation of the area, and by evaluating this expression at the bounds of -1 and 4, we found the precise numerical answer.

Function Inverses

Function inverses essentially reverse the roles of the input and output in a given function. If we have a function 'f' that takes 'x' to 'y', its inverse 'g' (if it exists) will take 'y' back to 'x'. The notation generally used for an inverse function is 'f^{-1}(x)'. However, not all functions are invertible — they need to be one-to-one and pass the Horizontal Line Test, which means that each 'y' value is unique to one 'x' value.

The inverse of a function is critical when we want to reverse the effect of the original function. In our exercise, we approximated the inverse function because finding an algebraic expression for the inverse was not straightforward. This approximation allowed us to proceed with finding the area under the curve, thus applying the concept of function inverses to a practical example in integral calculus.