Problem 15 - Calculus: Early Transcendentals

Question

Use Part 1 of the Fundamental Theorem of Calculus to find the derivative of the function. \( \displaystyle y = \int^{3x + 2}_1 \frac{t}{1 + t^3} \,dt \)

Step-by-Step Solution

Answer

The derivative is \( \frac{3(3x+2)}{1 + (3x+2)^3} \)."

Step 1: Understanding the Function

We are given the function \( y = \int^{3x + 2}_1 \frac{t}{1 + t^3} \,dt \) and asked to find its derivative with respect to \( x \). This is an application of the Fundamental Theorem of Calculus, specifically Part 1, which relates the derivative of an integral to its integrand.

Step 2: Identifying the Variable Limits

To use the Fundamental Theorem of Calculus Part 1, we need to identify the variable limit of integration. In the given integral, the upper limit \( 3x + 2 \) is a function of \( x \), and the lower limit is constant.

Step 3: Applying the Fundamental Theorem of Calculus Part 1

According to the Fundamental Theorem of Calculus Part 1, if \( F(x) = \int_{a}^{g(x)} f(t) \,dt \), then the derivative \( F'(x) = f(g(x)) \cdot g'(x) \). Here, \( f(t) = \frac{t}{1+t^3} \) and \( g(x) = 3x + 2 \).

Step 4: Calculating the Derivative

First, evaluate the integrand at the upper limit: \( f(3x+2) = \frac{3x+2}{1 + (3x+2)^3} \). Then, find the derivative of the upper limit: \( g'(x) = 3 \). Thus, \( y'(x) = \frac{3x+2}{1 + (3x+2)^3} \cdot 3 \).

Step 5: Simplifying the Expression

Multiply the expression to find the final derivative: \( y'(x) = 3 \cdot \frac{3x+2}{1 + (3x+2)^3} = \frac{3(3x+2)}{1 + (3x+2)^3} \).

Key Concepts

Derivative of an Integral

The Fundamental Theorem of Calculus connects differentiation and integration, two core concepts in calculus. Part 1, often called the "derivative of an integral," is key here. In its simplest terms, this part of the theorem states that if you have an integral like \[F(x) = \int_{a}^{g(x)} f(t) \, dt,\]where \(a\) is a constant and \(g(x)\) is a function of \(x\), then the derivative \(F'(x)\) can be calculated by evaluating the integrand \(f(t)\) at the upper limit \(g(x)\) and multiplying by the derivative of \(g(x)\). This gives us the formula:\[F'(x) = f(g(x)) \cdot g'(x).\]This rule allows us to quickly find the derivative of the integral without needing to solve the integral itself. Pretty nifty, right? Especially in cases where the integral could be very complex.

Variable Limits of Integration

When tackling an integral with variable limits, identifying which limit changes is crucial. For the expression \[y = \int^{3x + 2}_1 \frac{t}{1 + t^3} \,dt,\]observe that the upper limit, \(3x + 2\), is a function of \(x\), while the lower limit is a constant. The Fundamental Theorem of Calculus Part 1 handles variable limits by letting us differentiate with respect to the variable, here \(x\), by treating everything else as constant. This means we only focus on the upper limit's changes.
To differentiate properly, figure out:

• Which limit varies — in this case, the upper limit.
• How that limit depends on the variable — understand that it's \(3x + 2\), a linear function of \(x\).

This forms the groundwork for applying Part 1 of the Fundamental Theorem of Calculus.

Integrand Evaluation

Evaluating the integrand is the next step in using the Fundamental Theorem of Calculus Part 1. This involves substituting the variable limits back into the function, essentially replacing \(t\) with the upper limit of integration. For our example, the integrand is \[f(t) = \frac{t}{1+t^3}.\]Substitute \(t\) with \(3x + 2\) from the variable upper limit:\[f(3x+2) = \frac{3x+2}{1 + (3x+2)^3}.\]This substitution is important because it transforms the original function given under the integral into one explicit in terms of \(x\), the variable of differentiation.
Finally, we bring in the derivative of the upper limit, \(g(x)\), which is \(g'(x) = 3\). The integrand value multiplied by this derivative gives the final derivative result:\[y'(x) = \frac{3(3x+2)}{1 + (3x+2)^3}.\]Through these steps, the whole problem boils down to simple substitution and multiplication, thanks to this powerful theorem.