Problem 19 - Calculus

Question

Find \(G^{\prime}(x).\) $$ G(x)=\int_{0}^{x}\left(2 t^{2}+\sqrt{t}\right) d t $$

Step-by-Step Solution

Answer

\( G'(x) = 2x^2 + \sqrt{x} \)

Step 1: Identify the Type of Function

The given function \( G(x) = \int_{0}^{x} (2t^2 + \sqrt{t}) \, dt \) is defined as the integral of a function from a constant to \( x \). This type of function is known as a definite integral with a variable upper limit of integration.

Step 2: Apply the Fundamental Theorem of Calculus

According to the Fundamental Theorem of Calculus, if \( F(t) \) is the antiderivative of \( f(t) \), then the derivative of the integral of \( f(t) \) from a to \( x \) with respect to \( x \) is \( f(x) \). Therefore, \( G'(x) = 2x^2 + \sqrt{x} \).

Step 3: Simplify the Expression

The derivative \( G'(x) \) is already in simplest form, given by substituting \( t = x \) into \( f(t) = 2t^2 + \sqrt{t} \), resulting in \( G'(x) = 2x^2 + \sqrt{x} \).

Key Concepts

definite integral

A definite integral is a fundamental concept in calculus that offers a way to calculate the accumulation of quantities. When we talk about a definite integral from a lower limit to an upper limit, it generally gives us the total accumulated value between these two points. In the exercise, we have a function defined by a definite integral \( G(x) = \int_{0}^{x} (2t^2 + \sqrt{t}) \, dt \). This integral shows the accumulation of the function \( 2t^2 + \sqrt{t} \) from \( t = 0 \) to \( t = x \).

In practice, definite integrals are often used to find quantities such as areas under curves or total change over an interval. The result of a definite integral is a specific number when both the lower and upper limits are constants. However, in our case, the upper limit is a variable \( x \), making the integral a function of \( x \).

• The process involves finding the antiderivative of the integrand.
• Evaluating it at the upper limit and lower limit.
• Taking the difference to find the accumulated value.

antiderivative

An antiderivative represents a reverse operation to differentiation. It involves finding a function whose derivative produces the original function. In the context of the Fundamental Theorem of Calculus, understanding antiderivatives is essential.

The theorem states that if \( F(t) \) is an antiderivative of \( f(t) \), then the definite integral of \( f(t) \) over an interval can be evaluated using \( F(t) \)._An antiderivative of \( f(t) = 2t^2 + \sqrt{t} \) is \( F(t) \) such that \( F'(t) = 2t^2 + \sqrt{t} \)._ This helps establish the integral's value or function when evaluating between limits. Some important points include:

• The antiderivative is unique up to a constant.
• Finding antiderivatives requires the application of rules and techniques for reversing differentiation.

This allows calculation for varying limits as seen with the upper limit of integration being \( x \).

derivative of a function

Calculating the derivative of a function involves finding the rate at which the function value changes as its input changes. The derivative can provide insights like slope or velocity. For \( G(x) = \int_{0}^{x} (2t^2 + \sqrt{t}) \, dt \), finding \( G'(x) \) tells us about the behavior of the function \( G \) as \( x \) changes. By using the Fundamental Theorem of Calculus, we know:

\[ G'(x) = f(x) = 2x^2 + \sqrt{x} \]

This means that the rate of change of \( G(x) \) with respect to \( x \) is simply the original integrand evaluated at \( x \). Some key takeaways from derivatives include:

• They reveal instantaneous rates of change.
• Function behavior analysis like increasing or decreasing trends.
• Recognition of smooth curves, sharp points, or discontinuities.

The derivative simplifies the understanding of how an integral function changes over its interval, enhancing in-depth comprehension of accumulation and variation dynamics.