Question

Show that the value of \(\int_{0}^{1} \sin \left(x^{2}\right) d x\) cannot possibly be 2 .

Short answer

Question: Show that the value of \(\int_{0}^{1} \sin(x^2) dx\) cannot be 2. Answer: We find that the function \(\sin(x^2)\) is positive over the interval \([0,1]\) and has a maximum value of 1. The upper bound for the integral value is given by \(\int_{0}^{1} 1 dx\), which is equal to 1. Since the integral is definitely smaller than its upper bound, it cannot possibly be equal to 2. Thus, the value of \(\int_{0}^{1} \sin(x^2) dx\) cannot be 2.

Step-by-step solution

Analyze the function \(\sin(x^2)\)

The given function is \(\sin(x^2)\), which is a positive function on the interval \([0,1]\). Since \(\sin(x)\) has a maximum value of 1 and \(x^2\) is an increasing function on this interval, the maximum value of \(\sin(x^2)\) will also be 1.

Find the upper bound of the integral

To show that the value of the integral cannot possibly be 2, we need to look for an upper bound on its value. We know that \(\sin(x^2)\) has a maximum value of 1, and since it is positive in \([0,1]\), the largest possible value for the integral can be obtained when the function has the maximum value over the entire interval. In this case, the area under the curve would be equivalent to a rectangle with width 1 and height 1. The upper bound for the integral value is then: $$\int_{0}^{1} 1 dx$$

Evaluate the integral and compare with 2

Now, let's evaluate the integral to find the upper bound: $$\int_{0}^{1} 1 dx = x \Big|_{0}^{1} = 1 - 0 = 1$$ Since \(\int_{0}^{1} \sin(x^2) dx\) is definitely smaller than its upper bound, 1, it cannot possibly be equal to 2. Thus, we have shown that the value of \(\int_{0}^{1} \sin(x^2) dx\) cannot possibly be 2.

Key concepts

Definite Integrals

In integral calculus, a definite integral represents the area under a curve, within a specified range or interval. Defined by two boundaries, known as limits of integration, it measures the net accumulation of a function's values as the input changes across the range. For instance, when considering the integral \(\int_{a}^{b} f(x) dx\), 'a' and 'b' are the lower and upper bounds, respectively.

With the problem in hand, \(\int_{0}^{1} \sin(x^2) dx\), we are looking to find the area under the curve of \(\sin(x^2)\) from 0 to 1. Because the value of the sine function varies from -1 to 1, the area cannot exceed the bounds created by the rectangle with the interval [0,1] as its base and 1 as the height. By understanding definite integrals, we can confidently infer that the maximum area under our curve is bound by the value 1, and thus cannot be 2 as hypothesized in the exercise.

Integral Upper Bound

An integral's upper bound is critical for understanding and estimating the values of definite integrals. It is the upper limit up to which the area under the curve (integration) is calculated. In essence, it serves as a stop point for the aggregation of the function's values. However, it can also provide a 'maximum' estimation of the integral if the function's values stay below a certain limit within the integration interval.

In our example, \(\int_{0}^{1} \sin(x^2) dx\), the integral upper bound is 1. We know \(\sin(x^2)\) is less than or equal to 1 for any x within [0,1]. Therefore, if we replace \(\sin(x^2)\) with its maximum value of 1 for the same interval, we get \(\int_{0}^{1} 1 dx\), which gives us the upper bound estimate of the original integral. It is impossible for the original integral to exceed this upper bound, showcasing how integral calculus uses bounds to derive meaning from function behavior within specific intervals.

Properties of Sine Function

The sine function plays a pivotal role in trigonometry and calculus due to its widespread properties and applications. This periodic function is known for oscillating between -1 and 1. We observe that \(\sin(\theta)\) will always produce values within this range for any angle \(\theta\). Additionally, the sine function is continuous and smooth, making it a prime candidate for integration.

When the argument of the sine function is squared, as in \(\sin(x^2)\), its behavior remains confined within -1 and 1, although the rate of oscillation changes. This property ensures that for our problem, the integral of \(\sin(x^2)\) from 0 to 1 must be trapped within a predictable scope. This inherent nature of the sine function, along with its bounded output, is key to understanding the limitations of the definite integral presented in the exercise and further validates that the solution cannot exceed the upper bound of 1.