Problem 33 - Thomas Calculus

Question

In Exercises \(17-54\) , find the most general antiderivative or indefinite integral. Check your answers by differentiation. $$ \int \frac{t \sqrt{t}+\sqrt{t}}{t^{2}} d t $$

Step-by-Step Solution

Answer

The antiderivative is \(2t^{1/2} - 2t^{-1/2} + C\).

Step 1: Simplify the Integral

Start by simplifying the expression \( \frac{t \sqrt{t} + \sqrt{t}}{t^2} \). This can be rewritten as:\[ \int \left( \frac{t^{3/2}}{t^2} + \frac{t^{1/2}}{t^2} \right) \, dt \]This simplifies to:\[ \int \left( t^{-1/2} + t^{-3/2} \right) \, dt \]

Step 2: Integrate Term by Term

Now integrate each term separately. For \( \int t^{-1/2} \, dt \), use the power rule for integration:\[ \int t^n \, dt = \frac{t^{n+1}}{n+1} + C \] where \( n eq -1 \).Substitute \( n = -1/2 \):\[ \int t^{-1/2} \, dt = \frac{t^{1/2}}{1/2} = 2t^{1/2} \]For \( \int t^{-3/2} \, dt \):Substitute \( n = -3/2 \):\[ \int t^{-3/2} \, dt = \frac{t^{-1/2}}{-1/2} = -2t^{-1/2} \]

Step 3: Combine the Integrals

Combine the results of integrating each term:\[ 2t^{1/2} - 2t^{-1/2} + C \]This represents the most general antiderivative of the function.

Step 4: Verify by Differentiation

Differentiate the result \( 2t^{1/2} - 2t^{-1/2} + C \) to ensure it equals the original function.\[ \frac{d}{dt}(2t^{1/2}) = t^{-1/2} \]\[ \frac{d}{dt}(-2t^{-1/2}) = -(-1)t^{-3/2} = 2t^{-3/2} \]Combining these gives:\[ t^{-1/2} + t^{-3/2} \]Upon simplification, this matches the numerator of the original integral, confirming correct integration.

Key Concepts

Definite Integral

The concept of a definite integral is central in calculus and refers to the method of finding the exact area under a curve between two points on the x-axis. Unlike the indefinite integral, which finds a family of functions (antiderivatives), the definite integral computes a real number. It's used when specific interval boundaries are known.
To evaluate a definite integral, we apply the fundamental theorem of calculus. This process often involves finding the indefinite integral first, and then using the limits of integration to find the net area.

• The notation for a definite integral is \( \int_a^b f(x) \, dx \), where \( a \) and \( b \) are the limits of integration.
• Calculate the antiderivative \( F(x) \) of the function \( f(x) \).
• Evaluate \( F(b) - F(a) \) to get the value of the definite integral.

Power Rule

The Power Rule is an essential tool for integration and differentiation, offering a straightforward way to find antiderivatives and derivatives of polynomials. It simplifies the process by allowing you to work term by term.
For integration, the power rule states:

• If \( f(x) = x^n \), where \( n eq -1 \), then the integral \( \int x^n \, dx = \frac{x^{n+1}}{n+1} + C \), where \( C \) is the constant of integration.
• This formula is derived by doing the inverse of the differentiation power rule.

When integrating each term separately, like in the original example where powers are negative, the power rule still applies. You adjust \( n \) for each term and then apply the rule to find its antiderivative.

Differentiation

Differentiation is the process of computing the derivative, which shows how a function changes as its input changes. It plays a key role in calculus for verifying antiderivatives and ensuring correct integration.

• The derivative of a function represents the slope of its tangent line at any given point.
• It can be used to confirm the antiderivative by differentiating the result and checking if it matches the original function.
• The differentiation power rule states: \( \frac{d}{dx} x^n = nx^{n-1} \).

In the original solution, differentiating the antiderivative \( 2t^{1/2} - 2t^{-1/2} + C \) correctly verified it matched the original integrand, thus confirming the integration was done right.

Simplification of Expressions

Simplifying expressions is a key part of solving integrals and making them more manageable. Before integration, it's beneficial to rewrite complex expressions into simpler forms. This often involves breaking down fractions or using algebraic identities.

• In the original exercise, \( \frac{t \sqrt{t} + \sqrt{t}}{t^2} \) was simplified to \( t^{-1/2} + t^{-3/2} \).
• This was achieved by expressing \( \sqrt{t} \) as \( t^{1/2} \) and simplifying each component of the fraction separately.
• Simplification reduces complexity and aligns the expression with forms suitable for applying integration rules.

Such simplification allows for easier application of the Power Rule and enhances computation efficiency during integration.