Question

If \(g(x)=\int_{\sin x}^{\sin (2 x)} \sin ^{-1}(t) d t\), then [A] \(g^{\prime}\left(\frac{\pi}{2}\right)=-2 \pi\) [B] \(g^{\prime}\left(-\frac{\pi}{2}\right)=2 \pi\) [C] \(g^{\prime}\left(\frac{\pi}{2}\right)=2 \pi\) [D] \(g^{\prime}\left(-\frac{\pi}{2}\right)=-2 \pi\)

Short answer

None of the given options is correct, as \(g'(\frac{\pi}{2})\) and \(g'(-\frac{\pi}{2})\) both evaluate to 0.

Step-by-step solution

Use the Fundamental Theorem of Calculus Part 1

The problem involves finding the derivative of an integral with variable limits of integration. According to the Fundamental Theorem of Calculus Part 1, if we have an integral function defined by \(G(x) = \int_{a(x)}^{b(x)} f(t)\,dt\), the derivative \(G'(x)\) is \(f(b(x))b'(x) - f(a(x))a'(x)\). For the given problem, \(G(x) = g(x)\), \(a(x) = \sin(x)\), and \(b(x) = \sin(2x)\), the function \(f(t) = \sin^{-1}(t)\).

Differentiate the Upper and Lower Limits

To apply the theorem, differentiate the upper and lower limits with respect to \(x\). The derivatives are \(a'(x) = \cos(x)\) and \(b'(x) = 2\cos(2x)\).

Apply the Derivative of the Integral

Apply the derivative to the integral function using the derived expressions: \(g'(x) = f(b(x)) \cdot b'(x) - f(a(x)) \cdot a'(x) = \sin^{-1}(\sin(2x)) \cdot 2\cos(2x) - \sin^{-1}(\sin(x)) \cdot \cos(x)\).

Evaluate the Derivative at the Given Points

Evaluate \(g'(x)\) at the points \(x = \frac{\pi}{2}\) and \(x = -\frac{\pi}{2}\). At \(x = \frac{\pi}{2}\), \(g'\left(\frac{\pi}{2}\right) = \sin^{-1}(\sin(\pi)) \cdot 2\cos(\pi) - \sin^{-1}(\sin(\frac{\pi}{2})) \cdot \cos(\frac{\pi}{2}) = 0 \cdot (-2) - \frac{\pi}{2} \cdot 0 = 0\). At \(x = -\frac{\pi}{2}\), \(g'\left(-\frac{\pi}{2}\right) = \sin^{-1}(\sin(-\pi)) \cdot 2\cos(-\pi) - \sin^{-1}(\sin(-\frac{\pi}{2})) \cdot \cos(-\frac{\pi}{2}) = 0 \cdot (-2) - (-\frac{\pi}{2}) \cdot 0 = 0\). In both cases, the derivative evaluated at these points is 0, thus none of the given options [A], [B], [C], or [D] are correct.

Key concepts

Fundamental Theorem of Calculus

The Fundamental Theorem of Calculus serves as a bridge between integration and differentiation, two of the main operations in calculus. In its essence, this theorem connects the concept of the integral of a function to its antiderivative.

There are two parts to this theorem. The first part states that if a function is continuous on a closed interval \[a, b\], then the function has an antiderivative within this interval, and the integral of the function over \[a, b\] is the change in the values of this antiderivative. Symbolically, for a continuous function \(f\), if \(F\) is an antiderivative of \(f\), then: \[\int_{a}^{b} f(x)\text{ }dx = F(b) - F(a)\].

This theorem is a key tool in computing the definite integrals without the need for Riemann sums. In the context of the exercise given, the Fundamental Theorem of Calculus is applied to find the derivative of an integral with variable limits of integration, a concept which is elaborated in the following sections.

Derivative of an Integral

Understanding the derivative of an integral is crucial when dealing with variable limits of integration. According to the Fundamental Theorem of Calculus Part 1, to find the derivative of the integral function \( G(x) = \int_{a(x)}^{b(x)} f(t)\text{ }dt \), we use the expression \( G'(x) = f(b(x))\text{ }b'(x) - f(a(x))\text{ }a'(x) \).

This formula is applied whenever the limits of integration are functions of \( x \) instead of constants. In the exercise, the limits of \( g(x) \) are \( \sin(x) \) and \( \sin(2x) \) which makes the differentiation trickier than with constant limits. The ability to differentiate the variable limits is therefore as important as understanding the integral itself.

Variable Limits of Integration

When the limits of an integral are functions of the variable of differentiation, we refer to these as variable limits of integration. They add an additional layer of complexity because when differentiating, both the limits and the function inside the integral need to be considered.

For example, if we differentiate \( g(x) = \int_{a(x)}^{b(x)} f(t)\text{ }dt \), where \( a(x) \) and \( b(x) \) are functions of \( x \) rather than constants, we consider the rate at which these limits change. The derivatives \( a'(x) \) and \( b'(x) \) become part of our final expression when using the derivative of the integral formula described in the previous section. This attention to the variable limits is demonstrated in the exercise solution, where the derivatives of \( a(x) = \sin(x) \) and \( b(x) = \sin(2x) \) come into play.

Integration and Differentiation

Integration and differentiation are two fundamental operations in calculus, often seen as inverse processes. Differentiation is about finding the rate at which something changes, that is, finding the derivative. Integration, on the other hand, is the process of finding the quantity where the rate of change is known—this is equivalent to finding the antiderivative or the area under a curve.

While differentiation is commonly used to find the velocity of an object if its position as a function of time is known, integration can be used to find the position if the velocity is known. The tie between the two is cemented by the Fundamental Theorem of Calculus, which shows that integration can be undone by differentiation and vice versa when considering definite integrals. The exercise provided is an application of both operations, as it involves differentiating an integral where the integrand and the limits are functions of the variable.