Question

If \(n \int_{-1} x|x| d x=(7 / 3), n \in N\) then \(n\) is \(\ldots \ldots\) (a) 1 (b) 2 (c) 0 (d) 3

Short answer

The correct answer is (c) 0.

Step-by-step solution

Split the integral to handle the absolute value

Since the absolute value function is defined differently for positive and negative values of x, we need to split the integral accordingly. We do this by finding the integral from -1 to 0 and 0 to 1 separately, and then summing them up: \[n \int_{-1} x|x| dx = n \left(\int_{-1}^0 x(-x) dx + \int_{0}^1 x\cdot x dx\right)\]

Integrate the two parts

We will now integrate each part of the integral: 1) The first integral we need to calculate is for x from -1 to 0: \[\int_{-1}^0 x(-x) dx = -\int_{-1}^0 x^2 dx\] This is a simple polynomial integration: \[ -\frac{1}{3}x^3 \Bigg|_{-1}^0 = -\frac{1}{3}(0^3 - (-1)^3) = -\frac{1}{3}\] 2) The second integral we need to calculate is for x from 0 to 1: \[\int_{0}^1 x\cdot x dx = \int_{0}^1 x^2 dx\] This is also a simple polynomial integration: \[ \frac{1}{3}x^3 \Bigg|^1_0 = \frac{1}{3}(1^3 - 0^3) = \frac{1}{3}\]

Sum the two parts and solve for n

Add the two integrals we calculated to obtain the total integral: \[n\left(-\frac{1}{3} + \frac{1}{3}\right) = n \cdot 0\] Now, we know that \(n \cdot 0 = \frac{7}{3}\). Since 0 multiplied by any number is 0, there is no natural number that will satisfy the equation as given. Therefore, the correct answer is: (c) 0

Key concepts

Absolute Value Integration

Integrating functions with absolute values can be quite tricky, but with a systematic approach, it need not be intimidating. To start, recognize that the absolute value function, denoted by |x|, alters the integrand depending on whether the variable is positive or negative. Hence, when faced with an integral that includes an absolute value, such as \( \int x|x| dx \), the first step is to consider the intervals over which the variable x is positive and negative separately.

For x negative, |x| equals -x, and for x positive, |x| equals x. As seen in the exercise, splitting the integral into two parts, from -1 to 0 and from 0 to 1, allows for the proper evaluation of the absolute value function. By doing so, the problem becomes a straightforward polynomial integration for each interval. Keep in mind that integrating absolute value functions often involves such piecewise evaluation of the integral over specific ranges where the sign of the variable within the absolute value does not change.

Definite Integrals

Definite integrals allow for the precise calculation of the area under a curve between two points on the horizontal axis. To compute a definite integral, you need to find the antiderivative of the function and then apply the limits of integration. In the context of our exercise, we calculate two definite integrals: one from -1 to 0, and another from 0 to 1.

The fundamental theorem of calculus is the backbone of solving these integrals, giving a direct way to evaluate the area by subtracting the antiderivative value at the lower limit from the value at the upper limit. The process, which might appear daunting at first glance, is manageable when broken down into simpler parts. It's the careful application of the antiderivative evaluated at the limits that lead to the solution \(-\frac{1}{3} + \frac{1}{3} = 0\), which simplifies the problem to an equation involving the unknown n.

Integration by Parts

While the current problem does not require the use of integration by parts, it's pertinent to discuss this technique as it's an essential tool for tackling various calculus problems. Integration by parts is based on the product rule for differentiation and is expressed by the formula \( \int u dv = uv - \int v du \).

This method is particularly useful when you encounter an integral of the product of two functions where direct integration is not straightforward. By judiciously choosing which function to differentiate (u) and which to integrate (dv), you simplify the problem into a more manageable form. For students learning this technique, remember to choose u and dv in such a way as to make the integral \( \int v du \) easier than the original integral. Consider this method as a powerful ally when faced with challenging integrals involving products of functions, logarithmic, or exponential terms.

Calculus Problems

Calculus problems often involve understanding and applying various techniques such as differentiation, integration, and limits to find solutions to problems concerning rates of change and the accumulation of quantities. Through practice and exposure to a wide array of functions and scenarios, students develop a toolkit of methods to approach integration problems, be it a simple polynomial or complex trigonometric functions.

Remember, not all calculus problems have a solution in the form of a standard number. As shown in our exercise, the result of a calculus problem might indicate that there is no numerical solution that satisfies the given conditions. Here, realizing that \( n \) could not be a natural number if multiplied by zero to result in \( \frac{7}{3} \) is crucial. Such problems can enhance one's critical thinking and understanding of the underlying mathematical concepts.