Problem 8 - The Calculus with Analytic Geometry

Question

\(\int \frac{27 t^{3}-1}{\sqrt[3]{t}} d t\)

Step-by-Step Solution

Answer

81 / 11 t^{11 / 3}+ - 3/ 2 - c

Step 1 - Simplify the Integrand

Rephrase the integrand by expressing each term to a common base. Write \(t^{-1/3}\) for \(\frac{1}{\root 3 \t}\). Therefore, the integral becomes \(\begin{array}{l}\frac{27 t^{3}}{t^{1 / 3}}-\frac{1}{t^{1 / 3}}=\frac{27}{t} t^{3}-t^{-1 / 3} \ =27 t^{3-1/3}-t^{-1/3}=27 t^{8 / 3}-t^{-1/3}. \end{array}\)

Step 2 - Split the Integral

Separate the integral into two distinct terms: \(\begin{array}{l} \ =27 \int t^{8 / 3} d t-\frac{1}{\root 3 \t} d t.\end{array}\)

Step 3 - Integrate Each Part

Now integrate the two separated parts respectively. Recall the power rule for integration: \(\frac{t^{n+1}}{n+1}\). So, the integrals become \(\begin{array}{l}27 \frac{t^{11 / 3}}{11 / 3}- \frac{( \root 3 \t^{2})}{2 } =81 / 11 t^{11 / 3}-3/ 2 t^{ 5/3} + C \end{array}\)

Key Concepts

Integral Calculus

Integral calculus is all about finding the accumulation of quantities. It's like working backward from what you know about derivatives. Instead of finding the rate of change, you find the total quantity that makes up the change. Mathematically, integration helps you find areas under curves, among other things.

When dealing with integrals, you come across two types: definite and indefinite. The main difference is that definite integrals give you a specific number, while indefinite integrals give you a general equation with a constant of integration. Calculations you perform and the methods you use might slightly differ, but the core idea remains the same.

Power Rule for Integration

The power rule is a very convenient tool in integral calculus, especially when dealing with polynomials. It states that:
\ \ \( \int t^n dt = \frac{t^{n+1}}{n+1} + C \ \)

This rule is derived from reversing the power rule for differentiation. It's mostly used for any variable raised to a constant power. There are just a few points to remember:

• Don't forget to add 1 to the exponent.
• Divide by the new exponent.
• Always include the constant of integration, C, for indefinite integrals.
• This rule does not work for \( n = -1 \); for that, you use the natural logarithm instead.

Definite and Indefinite Integrals

In calculus, integrals can be classified into two main types:

• Indefinite Integrals: Represent a family of functions and include a constant of integration (C). They don't have limits of integration. For example, \( \int 27t^{8/3} dt \) results in a function plus C.
• Definite Integrals: Calculated over a specific interval \([a, b]\). They provide a specific value representing the area under the curve between \( a \) and \( b \). For definite integrals, you evaluate the antiderivative at the upper and lower bounds and subtract the values.

In our exercise, we simplified and split our integral to make use of indefinite integration, using the power rule to find solutions to each part.