Problem 25 - Calculus

Question

Find \(G^{\prime}(x).\) $$ G(x)=\int_{-x^{2}}^{x} \frac{t^{2}}{1+t^{2}} d t \text { Hint: } \int_{-x^{2}}^{x}=\int_{-x^{2}}^{0}+\int_{0}^{x} $$

Step-by-Step Solution

Answer

The derivative is \( G^{\prime}(x) = \frac{x^2}{1+x^2} + \frac{2x^5}{1+x^4}. \)

Step 1: Understand the Problem

We need to find the derivative of the function given as an integral with variable limits: \( G(x) = \int_{-x^2}^{x} \frac{t^2}{1+t^2} \, dt \). This requires us to apply the Fundamental Theorem of Calculus and Leibniz's Rule for differentiation under the integral sign.

Step 2: Split the Integral

As suggested by the hint, we can write the integral as the sum of two separate integrals: \[ G(x) = \int_{-x^2}^{0} \frac{t^2}{1+t^2} \, dt + \int_{0}^{x} \frac{t^2}{1+t^2} \, dt. \] This decomposition helps us apply the Fundamental Theorem of Calculus more conveniently.

Step 3: Differentiate the First Integral

For the integral \( \int_{-x^2}^{0} \frac{t^2}{1+t^2} \, dt \), apply Leibniz's rule which involves changing the upper limit and considering the effect of the lower limit. Using \( u = -x^2 \), the derivative becomes: \[ \text{For } F(t) = \frac{t^2}{1+t^2}, \quad \frac{d}{dx}\left(\int_{-x^2}^{0} F(t) \, dt\right) = -\frac{d}{du}(F(u)\cdot \frac{du}{dx}) = -\frac{d}{du}\left(\frac{(-x^2)^2}{1+(-x^2)^2}\right)(-2x). \] Evaluating this derivative gives: \[ \left(-1\right) \cdot \frac{2x \cdot \left(x^4/(1+x^4)\right)}{1+\left(-x^2\right)^2} = \frac{2x^5}{1+x^4}. \]

Step 4: Differentiate the Second Integral

For the integral \( \int_{0}^{x} \frac{t^2}{1+t^2} \, dt \), apply the Fundamental Theorem of Calculus: \[ \frac{d}{dx}\left(\int_{0}^{x} \frac{t^2}{1+t^2} \, dt\right) = \frac{x^2}{1+x^2}. \] This is straightforward since the upper limit is a function of \( x \) and the lower limit is a constant.

Step 5: Combine the Results

Add the results from the derivatives of both integrals:\[ G^{\prime}(x) = \frac{x^2}{1+x^2} + \frac{2x^5}{1+x^4}. \] Simplify this expression if needed for final clarity.

Key Concepts

Leibniz's Rule

Leibniz's Rule is a useful concept in calculus that allows us to differentiate an integral where the limits of integration are both functions of a variable, usually denoted as \( x \). This is handy in solving problems involving integrals with variable limits. The rule states that if you have an integral of the form:

• \( rac{d}{dx} \left( \int_{u(x)}^{v(x)} f(t) \, dt \right) = f(v(x))v'(x) - f(u(x))u'(x) + \int_{u(x)}^{v(x)} \frac{\partial}{\partial x} f(t, x) \, dt \)

Here, \( f(t,x) \) could be a function involving \( t \) and \( x \), and \( u(x) \) and \( v(x) \) are the variable limits. Leibniz's Rule effectively combines the Fundamental Theorem of Calculus with differentiation techniques to handle integrals with dynamic boundaries.
In our exercise, we use this rule to manage how changes in \( x \) affect the integral \( \int_{-x^2}^{0} \frac{t^2}{1+t^2} \, dt \) as \( x \) changes. This application involves recognizing that both the upper and lower limits are dependent on \( x \). Understanding that the derivative of \( u(x) = -x^2 \) is \(-2x\) is key to finding the overall derivative using this rule.

Differentiation under integral sign

Differentiation under the integral sign is a technique which is part of more general solutions like Leibniz's Rule. This method is utilized when we need to differentiate an integral whose integrand includes parameters depending on the differentiation variable. It allows for differentiation directly by treating other variables as constants within the integral.
In the context of our exercise, finding \( G'(x) \) for \( G(x) = \int_{0}^{x} \frac{t^2}{1+t^2} \, dt + \int_{-x^2}^{0} \frac{t^2}{1+t^2} \, dt \) involved differentiating under the integral sign. The second part was straightforward. We simply applied the Fundamental Theorem of Calculus to the integral from \(0\) to \(x\), as it has a constant lower limit and a differentiable upper limit in terms of \(x\).

• The result was \( \frac{d}{dx}(\int_{0}^{x} \frac{t^2}{1+t^2} \, dt) = \frac{x^2}{1+x^2} \).

This technique shows the power of differentiation under the integral sign, allowing for simplification and direct calculation of the result.

Variable limits of integration

Variable limits of integration occur when one or both of the limits in a definite integral are themselves variables, rather than constants. This requires special consideration when taking derivatives. The limits depend on the variable with respect to which differentiation is performed.
In the given problem, we have a function \( G(x) \) defined as an integral with limits that are variable and depend on \( x \):

• \( G(x) = \int_{-x^2}^{x} \frac{t^2}{1+t^2} \, dt \)

Here, the lower limit \(-x^2\) is a function of \( x \), and the upper limit is simply \( x \). This requires the use of techniques like Leibniz's Rule to differentiate correctly, accounting for how each limit contributes to the derivative.
Using variable limits makes problems more complex, as seen in the exercise: applying Leibniz’s Rule and the Fundamental Theorem allowed us to systematically differentiate these integrals. We split the integral at zero and found the contribution from each part separately.