Problem 9 - Calculus Early Transcendentals

Question

Evaluate \(\frac{d}{d x} \int_{a}^{x} f(t) d t\) and \(\frac{d}{d x} \int_{a}^{b} f(t) d t,\) where \(a\) and \(b\) are constants.

Step-by-Step Solution

Answer

1. \(\frac{d}{dx} \int_{a}^{x} f(t) dt\) 2. \(\frac{d}{dx} \int_{a}^{b} f(t) dt\) Answer: The derivatives of the given definite integrals are: 1. \(\frac{d}{dx} \int_{a}^{x} f(t) dt = f(x)\) 2. \(\frac{d}{dx} \int_{a}^{b} f(t) dt = 0\)

Step 1: Fundamental Theorem of Calculus

The first part of the Fundamental Theorem of Calculus states that if \(F(x)\) is the antiderivative of \(f(x)\), then: $$\int_{a}^{x} f(t) dt = F(x) - F(a)$$ Knowing this, we can directly apply the theorem to compute the derivatives.

Step 2: Derivative of the first integral

First, we calculate the derivative of the integral with respect to x: $$\frac{d}{dx} \int_{a}^{x} f(t) dt$$ Applying the Fundamental Theorem of Calculus, we get: $$\frac{d}{dx}(F(x)-F(a))$$ Since \(F(a)\) is a constant, its derivative with respect to \(x\) is \(0\). Therefore we are left with: $$\frac{d}{dx} F(x)$$ Since \(F(x)\) is the antiderivative of \(f(x)\), taking its derivative with respect to \(x\) gives us back the function \(f(x)\): $$\frac{d}{dx} \int_{a}^{x} f(t) dt = f(x)$$

Step 3: Derivative of the second integral

Now, we calculate the derivative of the second integral with respect to \(x\): $$\frac{d}{dx} \int_{a}^{b} f(t) dt$$ We notice that both limits of integration, \(a\) and \(b\), are constants. This means that the integral itself is a constant. Let's call this constant \(C\): $$C = \int_{a}^{b} f(t) dt$$ Now, we calculate the derivative with respect to \(x\): $$\frac{d}{dx} C$$ Since the constant \(C\) has no dependence on \(x\), its derivative with respect to \(x\) is zero: $$\frac{d}{dx} \int_{a}^{b} f(t) dt = 0$$

Step 4: Final Results

We have found the derivatives of the given integrals: 1. \(\frac{d}{dx} \int_{a}^{x} f(t) dt = f(x)\) 2. \(\frac{d}{dx} \int_{a}^{b} f(t) dt = 0\)