Problem 67 - Biocalculus Calculus for the Life Sciences

Question

\(67-68\) Sketch the area represented by \(g(x) .\) Then find \(g^{\prime}(x)\) in two ways: (a) by using Part 1 of the Fundamental Theorem and (b) by evaluating the integral using Part 2 and then differentiating. \(g(x)=\int_{0}^{x}\left(1+t^{2}\right) d t\)

Step-by-Step Solution

Answer

The derivative \(g'(x)\) is \(1 + x^2\).

Step 1: Sketch the Integral Function Graphically

To sketch the area represented by the given function \(g(x)\), we need to consider the integrand \(1 + t^2\). This expression denotes a parabola opening upwards. The integral from 0 to \(x\) accumulates the area under this curve from \(t = 0\) to \(t = x\). Graphically, this means that you consider the area under the curve \(1 + t^2\) starting from the y-axis and ending at the vertical line \(t=x\).

Step 2: Apply Part 1 of the Fundamental Theorem of Calculus

Part 1 of the Fundamental Theorem of Calculus tells us that if \(g(x) = \int_{a}^{x} f(t) \, dt\), then \(g'(x) = f(x)\). Here, \(f(t) = 1 + t^2\). Therefore, by applying this theorem, we find:\[g'(x) = 1 + x^2 .\]

Step 3: Evaluate the Integral Using Part 2 of the Fundamental Theorem of Calculus

The Part 2 of the Fundamental Theorem of Calculus states that if \(F\) is an antiderivative of \(f\) on \([a, b]\), then \(\int_{a}^{b} f(t) \, dt = F(b) - F(a)\). Here, \(F(t)\) that is an antiderivative of \(f(t) = 1 + t^2\) is \(t + \frac{t^3}{3}\).

Step 4: Find the Antiderivative and Evaluate the Integral

Evaluate the integral \(g(x) = \int_{0}^{x} (1 + t^2) \, dt\) by finding the antiderivative:\[F(t) = t + \frac{t^3}{3}.\]Now, evaluate \(F(x) - F(0)\):\[g(x) = \left(x + \frac{x^3}{3}\right) - \left(0 + \frac{0^3}{3}\right) = x + \frac{x^3}{3}.\]

Step 5: Differentiate the Resulting Expression

Differentiate the resulting expression from Step 4 to find \(g'(x)\):\[g'(x) = \frac{d}{dx}\left(x + \frac{x^3}{3}\right) = 1 + x^2.\]

Step 6: Conclusion: Verify the Expression

Both methods, Part 1 and Part 2 of the Fundamental Theorem of Calculus, yield the same derivative \(g'(x) = 1 + x^2\), confirming the correctness of the solution.

Key Concepts

Definite Integrals

The concept of definite integrals involves calculating the area under a curve defined by a function over a specific interval. It's like collecting small strips of area and adding them up. In this exercise, the function is given as \(g(x) = \int_{0}^{x}(1+t^2 ) dt\).

• The limits of integration, from 0 to \(x\), specify where the calculation starts and stops.
• The integrand, \(1+t^2\), tells us the shape of the curve we're measuring the area under.

To visualize this, imagine drawing a curve \(y = 1 + t^2\) and shading the area from \(t = 0\) to \(t = x\). The definite integral captures the amount of space beneath the curve within these bounds. This approach is an essential part of integral calculus.

Antiderivatives

Antiderivatives, sometimes known as indefinite integrals, are functions that represent the reverse process of taking a derivative. For example, if you know that \(f(t) = 1 + t^2\), then the antiderivative \(F(t)\) would be a function that differentiates back to \(f(t)\).

• In this problem, \(F(t) = t + \frac{t^3}{3}\) is the antiderivative of \(1 + t^2\).
• An antiderivative adds a consistent constant of integration, but for definite integrals, you subtract results at different bounds, so it cancels out.

Knowing antiderivatives allows us to evaluate definite integrals by applying the Fundamental Theorem of Calculus, which directly uses these concepts to compute areas under curves.

Differentiation

Differentiation refers to the process of finding the derivative of a function, which tells us how the function changes as its input changes. Derivatives represent rates of change or slopes of tangent lines on curves.

• For the function \(g(x) = x + \frac{x^3}{3}\), we differentiate to find \(g'(x)\).
• Applying differentiation gives \(g'(x) = \frac{d}{dx}(x + \frac{x^3}{3}) = 1 + x^2\).

This step confirms the rate of change of the accumulated area under the curve described by our original function. Differentiation is heavily used in integral calculus to verify the results of definite integrals and confirm their correctness.

Integral Calculus

Integral calculus involves the concept of integration, which is used to find areas, volumes, central points, among other things. It's essentially the reverse process of differentiation. Integral calculus deals with two primary operations:

• Finding the antiderivative (indefinite integration).
• Computing the definite integral (determining the actual area under a curve).

This problem uses both parts of the Fundamental Theorem of Calculus, which connects differentiation and integration. Part 1 allows immediate computation of a derivative from an integral, while Part 2 offers a method to evaluate a definite integral using antiderivatives. Together, they form the backbone of integral calculus, linking the geometry of shapes under curves to their algebraic expressions.