Question

Evaluate the following definite integrals. If \(\int_{0}^{k}(6 x-1) d x=4,\) find \(k\).

Short answer

Answer: The value of \(k\) that solves the definite integral is \(k \approx 1.5\).

Step-by-step solution

Find the antiderivative of \((6x-1)\)

To find the antiderivative of \((6x-1)\), we'll integrate the function with respect to \(x\). Remember that when integrating a polynomial, we increase the exponent by 1 and divide by the new exponent. The antiderivative of \((6x-1)\) is: $$ \int(6x-1)dx = 6\int x dx - \int 1 dx = 6\frac{x^2}{2} - x + C, $$ which simplifies to: $$ 3x^2 - x + C. $$

Apply the Fundamental Theorem of Calculus

The Fundamental Theorem of Calculus states that for a continuous function \(f(x)\) on the interval \([a, b]\), the definite integral between those two points is equal to the difference between the antiderivatives at those two points. In this case, we have: $$ \int_{0}^{k}(6 x-1) dx = (3k^2 - k) - (3(0)^2 - 0). $$ Simplifying the equation, we get: $$ \int_{0}^{k}(6 x-1) dx = 3k^2 - k. $$

Solve for k

We can now use the given condition to solve for \(k\). We know that \(\int_{0}^{k}(6 x-1) dx = 4\), so we can set up our equation: $$ 3k^2 - k = 4. $$ To solve for \(k\), first move the constant term to the left-hand side of the equation to get: $$ 3k^2 - k - 4 = 0. $$ Now, we want to solve this quadratic equation. Since it doesn't easily factor, we can apply the quadratic formula: $$ k = \frac{-b \pm \sqrt{b^2 - 4ac}}{2a}. $$ In our case, \(a=3\), \(b=-1\), and \(c=-4\). Plugging these values in, we get: $$ k = \frac{1 \pm \sqrt{(-1)^2 - 4 (3)(-4)}}{2 (3)}. $$ Solving further, we obtain: $$ k = \frac{1 \pm \sqrt{49}}{6}. $$ This gives us two possible values for \(k\): $$ k_1 = \frac{1 + \sqrt{49}}{6} \approx 1.5, $$ and $$ k_2 = \frac{1 - \sqrt{49}}{6} \approx -1.33. $$ Since the integral's upper limit of integration is greater than the lower limit, \(k_1 = 1.5\) is the correct answer.

Key concepts

Definite Integral

In calculus, a definite integral is fundamental to finding the area under a curve within a specific interval on the x-axis. Imagine it as the accumulation of infinite tiny rectangles under the curve between two points, say, \(a\) and \(b\) on the x-axis. The

• lower limit \(a\) is where you start.
• upper limit \(b\) is where you end.

Signified as \(\int_{a}^{b} f(x) \, dx\), the definite integral gives a numerical value representing either the area, volume, or other accumulative effects depending on the context.
In our exercise, we're evaluating \(\int_{0}^{k} (6x - 1) \, dx\), where the limits are from 0 to \(k\), and the result is given to be 4, helping us find the value of \(k\).

Antiderivative

An antiderivative of a function is another function whose derivative equals the original function. Often referred to as the "indefinite integral," finding an antiderivative is the reverse process of differentiation.
Consider the function \(6x - 1\). Its antiderivative is found by integrating it:

• Increase each exponent by 1.
• Divide by the new exponent.

For example:\[\int (6x - 1) \, dx = 6 \int x \, dx - \int 1 \, dx = 3x^2 - x + C,\]where \(C\) is the constant of integration.
This process is fundamental to solving definite integrals as it sets up the function needed for further evaluation.

Fundamental Theorem of Calculus

The Fundamental Theorem of Calculus cleverly unites differentiation and integration, forming a bridge between both concepts.
It states that if \(f(x)\) is continuous over an interval \([a, b]\) and \(F(x)\) is an antiderivative of \(f(x)\), then:\[\int_{a}^{b} f(x) \, dx = F(b) - F(a).\]In our exercise, this theorem is applied as follows:\[\int_{0}^{k} (6x - 1) \, dx = [3k^2 - k] - [3 \times 0^2 - 0] = 3k^2 - k.\]This shows how integrating over an interval yields the difference between the antiderivative evaluated at the two endpoints. It reduces the task of finding definite integrals to simply evaluating a function at two points.

Quadratic Equation Solving

A quadratic equation takes the general form \(ax^2 + bx + c = 0\). Solving it often involves factoring, using the quadratic formula, or completing the square. In our scenario:

• The result from solving the definite integral gives us the equation \(3k^2 - k = 4\).
• To further solve this, rearrange it to standard form: \(3k^2 - k - 4 = 0\).
• Apply the quadratic formula: \(k = \frac{-b \pm \sqrt{b^2 - 4ac}}{2a} \).

Using \(a = 3\), \(b = -1\), and \(c = -4\), the solution to our quadratic equation is:\[k = \frac{1 \pm \sqrt{49}}{6},\]giving us \(k_1 = 1.5\) and \(k_2 = -1.33\).
The correct \(k\) is \(k_1 = 1.5\) since it logically fits our problem's context. Quadratic equations can often provide multiple solutions, so it's critical to confirm which solution is appropriate based on the problem.