Problem 43 - University Calculus: Early Transcendentals

Question

Find the derivatives. a. by evaluating the integral and differentiating the result. b. by differentiating the integral directly. $$\frac{d}{d x} \int_{0}^{x^{3}} e^{-t} d t$$

Step-by-Step Solution

Answer

The derivative is \( 3x^2 e^{-x^3} \).

Step 1: Calculate the Integral

Let's first evaluate the integral \( \int_{0}^{x^3} e^{-t} \, dt \). Let \( u = x^3 \), so the integral becomes \( F(u) - F(0) \), where \( F(t) \) is the antiderivative of \( e^{-t} \). The antiderivative is \( -e^{-t} \), hence:\[ \int_{0}^{x^3} e^{-t} \, dt = -e^{-x^3} + e^0 = 1 - e^{-x^3} \]

Step 2: Differentiate the Resulting Function

Now, we differentiate the result of the integral with respect to \( x \):\[ \frac{d}{dx}(1 - e^{-x^3}) = 0 + 3x^2 e^{-x^3} \] This makes use of the chain rule for \( e^{-x^3} \), where the derivative of \(-x^3\) is \(-3x^2\).

Step 3: Apply the Fundamental Theorem of Calculus

Apply part 1 of the Fundamental Theorem of Calculus to differentiate the integral directly:Since the upper limit is a function of \( x \), we apply the chain rule, which gives us:\[ \frac{d}{dx} \int_{0}^{x^3} e^{-t} \, dt = e^{-x^3} \cdot \frac{d}{dx}(x^3) = e^{-x^3} \cdot 3x^2 = 3x^2 e^{-x^3} \]

Step 4: Confirm Both Methods Yield the Same Result

Both methods, evaluating and differentiating the integral and applying the Fundamental Theorem of Calculus directly, give the derivative:\[ 3x^2 e^{-x^3} \]

Key Concepts

chain rule in calculus

The chain rule in calculus is a fundamental technique used to differentiate composite functions. Whenever the argument of a function is itself another function, we use the chain rule. For instance, when differentiating something like \( e^{-x^3} \), it's essential to identify the inner and outer functions.
The outer function here is \( e^{-x^3} \), and the inner function is \( -x^3 \). By applying the chain rule, we first differentiate the outer function with respect to the inner function, and then multiply by the derivative of the inner function.
This gives us the derivative \( 3x^2 e^{-x^3} \):

• Differential of the outer function: \( e^{-x^3} \) becomes (keeping the exponent the same): \( -e^{-x^3} \).
• Differential of the inner function: \(-x^3\) becomes \(-3x^2\).
• Multiply these together: \(-3x^2 \cdot -e^{-x^3} = 3x^2 e^{-x^3}\).

Using the chain rule correctly simplifies the process of differentiating complex functions inside integrals or combined with exponentials.

evaluating definite integrals

Evaluating definite integrals involves finding the exact area under a curve defined by a function over a specific interval. To solve \( \int_{0}^{x^3} e^{-t} \, dt \), follow these steps:

• First, determine the antiderivative of the function. Here, we need the antiderivative of \( e^{-t} \), which is \( -e^{-t} \).
• Apply the limits of integration to this antiderivative. We have \( F(x^3) - F(0) \).
• Substitute the upper limit and the lower limit into the antiderivative to get the evaluated integral: \[ -e^{-x^3} + e^0 = 1 - e^{-x^3} \]

This completed result tells us how the accumulation of our function \( e^{-t} \) between 0 and \( x^3 \) behaves. It prepares us for further differentiation if needed.

differentiation of integrals

The differentiation of integrals is often approached using the Fundamental Theorem of Calculus. This theorem states that if you have a continuous function and its antiderivative, you can differentiate it directly using certain rules.
In this problem, we look at the integral \( \int_{0}^{x^3} e^{-t} \, dt \) with respect to \( x \). By the Fundamental Theorem, differentiate the integral directly if the limits of integration depend on \( x \):

• The derivative of \(\int_{a}^{u(x)} f(t) \, dt\) is \( f(u(x)) \cdot u'(x) \).
• In our exercise, apply the function \( e^{-x^3} \) and multiply by the derivative of the upper limit \( x^3 \), which is \( 3x^2 \).

Thus, differentiating the integral directly gives us \( 3x^2 e^{-x^3} \), confirming the derivative we evaluated previously. This method bypasses evaluating the integral first, providing a more immediate result leveraging calculus's powerful connections.