Question

Give a geometric explanation to explain why \(\int_{0}^{\pi / 2} x \sin x d x \leq \int_{0}^{\pi / 2} x d x\) Verify the inequality by evaluating the integrals.

Short answer

Geometrically, for the given range, the area under the function \(x \sin x\) will be smaller than the area under the function \( x \), since \( \sin x\) is always less than or equal to 1. Direct evaluation of the integrals also supports the statement with \( \pi/2 - 1 < \frac{\pi^2}{8} \).

Step-by-step solution

Geometric Explanation

A definite integral like \( \int_{0}^{\pi / 2} f(x) dx \) represents the area under the curve \( f(x) \) from \( x=0 \) to \( x=\frac{\pi}{2} \). In this case, \( x \sin x \) is less than or equal to \( x \) for \( x \) in the interval \([0, \pi/2]\), since \( \sin x \) is less than or equal to 1. Thus the area under the curve of \( x \sin x \) is less than or equal to the area under the curve of \( x \), over the interval \([0, \pi/2]\). This explains why \( \int_{0}^{\pi / 2} x \sin x dx \leq \int_{0}^{\pi / 2} x dx \).

Evaluate Integral of \( x \)

Firstly, evaluate \( \int_{0}^{\pi / 2} x dx \). The integral of \( x \) with respect to \( x \) is \( \frac{1}{2}x^2 \). Evaluated from \( x=0 \) to \( x=\pi/2 \), we get \( \frac{1}{2}(\pi/2)^2 = \frac{\pi^2}{8} \).

Evaluate Integral of \(x \sin x \)

Next, evaluate \( \int_{0}^{\pi / 2} x \sin x dx \). This can be computed using integration by parts with \( u=x \), \( dv=\sin x dx \), then \( du=dx \) and \( v=-\cos x \). Integration by parts formula is \( \int udv = uv - \int vdu \). Apply the formula and evaluate the resultant integrated terms from 0 to \( \pi/2 \). We would get \( [\pi/2] - 1 = \pi/2 - 1 \).

Compare The Evaluated Integrals

Finally, we'd compare the two evaluated integrals from Step-2 and Step-3. In our case, \( \pi/2 - 1 \) is indeed less than \( \frac{\pi^2}{8} \), which confirms our inequality.

Key concepts

Geometric Interpretation of Integrals

The definite integral represents the total area under a curve within a specified range. To visualize this, imagine the graph of a function on the x-y plane. The region between the function's graph and the x-axis, from one x-value to another, is the area the definite integral quantifies.

Consider the functions in the provided exercise, where we compare the areas under two curves: one is the curve of the function \( x \sin x \) and the other is for the function \( x \). Since the sine function oscillates between -1 and 1, and we are looking in the interval from 0 to \( \frac{\pi}{2} \), \( \sin x \) remains non-negative and never exceeds 1. Hence, the product \( x \sin x \) will be less or equal to \( x \) for all \( x \) in this interval. The geometric interpretation can be seen by actually sketching the two functions and literally seeing that the area under the curve \( x \sin x \) is lesser than the one under the curve \( x \), for \( x \) in \( [0, \pi/2] \).

Integration by Parts

Integration by parts is a technique which is particularly useful when dealing with the product of functions. It follows from the product rule for differentiation and is given by the formula: \[ \int u dv = uv - \int v du \].

In other words, choose \( u \) and \( dv \) such that the integral of \( v du \) (after finding \( v \) by integrating \( dv \)) is easier to evaluate than the original integral. Applying the technique to the function \( x \sin x \), we select \( u = x \) and \( dv = \sin x dx \). After differentiation and integration, we apply the formula and simplify to find the area under the curve.

Integral Evaluation

Evaluating integrals, essentially, means to find the exact area under the curve for the given function and within the specified limits. For the integral \( \int_{0}^{\pi / 2} x dx \), we find the antiderivative of \( x \), which is \( \frac{1}{2}x^2 \), and then apply the limits of integration to obtain its numerical value.

Similarly, for the integral \( \int_{0}^{\pi / 2} x \sin x dx \), after applying integration by parts, we find an expression for the area that we can evaluate at the bounds \( 0 \) and \( \pi/2 \). This process proves the established inequality algebraically by showing one integral is indeed lesser in value than the other.

Calculus Inequalities

Calculus inequalities are an important aspect of mathematical analysis that allow comparisons of the sizes of the values that functions can take. When it comes to definite integrals, we often use inequalities to compare the cumulative effects of functions over an interval.

In the given exercise, we establish an inequality between two definite integrals by considering the functions they integrate and the interval over which they integrate. It's an application of knowledge of function behaviors, like the fact that \( \sin x \) doesn't exceed 1, combined with the geometric interpretation and integration techniques. Such inequalities not only reassure our understanding of function behavior but also have practical implications, like in calculating areas and volumes in different fields like physics and engineering.