Question

Find the positive value of $$k$$ such that the area of the region nclosed between the graph of $$y=k \cos x$$ and the graph of $$y=k x^{2}$$ is $$2 .

Short answer

The value of \(k\) making the area of the region enclosed between the graph of \(y=k\cos x\) and the graph of \(y=kx^{2}\) equal to 2 can be calculated by solving the given equations and integral as described in the step-by-step solution.

Step-by-step solution

Setup equation

We'll start by setting up an equation that represents the area under the curve. We have the area \(A = \int_{a}^{b} |f(x) - g(x)| dx\) where \(f(x)\) and \(g(x)\) are our given functions. Since we know the area is equal to 2, our equation will be \(2 = \int_{a}^{b} |k\cos x - kx^{2}| dx\) where \(a\) and \(b\) are the x-coordinates at which the graphs of \(k\cos x\) and \(kx^{2}\) intersect.

Find intersection points

To find \(a\) and \(b\), solve for \(x\) in the equation \(k\cos x = kx^{2}\) yielding \(x=0\) and \(x=\sqrt{\cos x}\). In the positive region, we'll consider \(a=0\) and \(b=\sqrt{\cos a}\). So, our equation now becomes \(2 = \int_{a}^{b} |k\cos x - kx^{2}| dx\)

Solve the integral

Now we have to solve for \(k\) in the integral equation. Note that \(k\) is a factor in both functions and can be factored out. Also, given that k is positive, we can drop the absolute value bars around the integral, providing \(2 = |k|*\int_{0}^{b} (\cos x - x^{2}) dx\). Solve this integral and rearrange for \(k\).

Calculate the result

After finding the integral, divide 2 by the result to get \(k\). If the calculated \(k\) is negative, take absolute value since \(k\) is given to be positive.

Key concepts

Definite Integral

A crucial concept in calculus is the definite integral, which provides a way to calculate the accumulated value of a function over a specific interval. The definite integral is typically represented as \( \int_{a}^{b} f(x)\,dx \), where \( a \) and \( b \) denote the lower and upper bounds of the interval, and \( f(x) \) is the function being integrated. When applied to the problem of finding the area between curves, the definite integral measures the total area accumulated by the difference between the top curve and the bottom curve over the interval from \( a \) to \( b \).

In the context of the provided exercise, the definite integral is used to represent the area between the curves \( y = k\cos x \) and \( y = kx^{2} \). Since the task is to find the value of \( k \) that makes this area equal to 2, the equation becomes \( 2 = \int_{a}^{b} |k\cos x - kx^{2}|\,dx \), emphasizing that the definite integral is the key tool for computing the desired area. Here, the integral's bounds \( a \) and \( b \) will be the points where the functions intersect, which we will address in the next section.

Function Intersection Points

The term function intersection points refers to the x-values where two functions, \( f(x) \) and \( g(x) \), have the same y-value, meaning the graphs intersect. Finding the intersection points is essential when calculating the area between curves since these points typically serve as the bounds in the definite integral.

In our exercise, we need to determine the intersection points for \( y = k\cos x \) and \( y = kx^{2} \). When the functions are set equal to each other \( k\cos x = kx^{2} \), the k-values cancel out since they are constant multipliers and thus do not affect the x-values at which the intersections occur. Solving \( \cos x = x^{2} \) allows us to find the necessary bounds. For this case, we consider the positive roots because we're looking for a positive value of \( k \). As mentioned in the solution steps, the intersection points are \( x = 0 \) and \( x = \sqrt{\cos x} \), which give us the bounds for the integral necessary to solve for \( k \).

Solving Integrals in Calculus

When it comes to solving integrals in calculus, there are various techniques that can be applied depending on the type of function integrated. For the area between curves, the integral can often be simplified by factoring out constants or applying properties such as symmetry.

To solve the integral in the exercise, we start by factoring out the constant \( k \) because it is present in both terms of the integrand \( |k\cos x - kx^{2}| \). This simplification is significant because, after doing so, we are able to focus on the integral of the expression \( \cos x - x^{2} \) alone. Integrating this expression over the interval from \( a \) to \( b \), we obtain a result in terms of \( x \), which we can then evaluate at the bounds. Dividing 2 by this result will yield the positive value of \( k \).

It's also essential to understand that when integrating absolute value functions, as in \( |k\cos x - kx^{2}| \), one has to consider intervals where the function inside the absolute value changes sign. This nuance is key for solving the integral correctly and can require splitting the integral into multiple parts where the function maintains a consistent sign.