Table of contents

0. Functions7h 52m
1. Limits and Continuity2h 2m
2. Intro to Derivatives1h 33m
3. Techniques of Differentiation3h 18m
4. Applications of Derivatives2h 38m
5. Graphical Applications of Derivatives6h 2m
6. Derivatives of Inverse, Exponential, & Logarithmic Functions2h 37m
7. Antiderivatives & Indefinite Integrals1h 26m
8. Definite Integrals4h 44m
9. Graphical Applications of Integrals2h 27m
- Area Between Curves
  1h 18m
- Introduction to Volume & Disk Method
  1h 8m
10. Physics Applications of Integrals 3h 16m
- Kinematics
  1h 13m
- Work
  2h 3m
11. Integrals of Inverse, Exponential, & Logarithmic Functions2h 34m
12. Techniques of Integration7h 39m
13. Intro to Differential Equations2h 55m
14. Sequences & Series5h 36m
15. Power Series2h 19m
- Introduction to Power Series
  1h 9m
- Taylor Series & Taylor Polynomials
  1h 10m
16. Parametric Equations & Polar Coordinates7h 58m

0. Functions

Introduction to Functions

Calculus

0. Functions

Introduction to Functions

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Multiple Choice

Evaluate the definite integral: $\int_{0}^{1} \frac{2}{4} t^{3} - \frac{3}{4} t^{2} + \frac{2}{5} t d t$ from $t = 0$ to $t = 1$ .

\frac{3}{10}

\frac{1}{4}

0

\frac{1}{10}

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Verified step by step guidance

Step 1: Begin by identifying the integral to evaluate: $ \int_{0}^{1} \left( \frac{2}{4}t^3 - \frac{3}{4}t^2 + \frac{2}{5}t \right) \, dt $. Simplify the coefficients where possible, such as $ \frac{2}{4} = \frac{1}{2} $. The integral becomes $ \int_{0}^{1} \left( \frac{1}{2}t^3 - \frac{3}{4}t^2 + \frac{2}{5}t \right) \, dt $.

Step 2: Apply the power rule for integration to each term. Recall that the power rule states $ \int t^n \, dt = \frac{t^{n+1}}{n+1} $. For each term: $ \frac{1}{2}t^3 $ becomes $ \frac{1}{2} \cdot \frac{t^4}{4} $, $ -\frac{3}{4}t^2 $ becomes $ -\frac{3}{4} \cdot \frac{t^3}{3} $, and $ \frac{2}{5}t $ becomes $ \frac{2}{5} \cdot \frac{t^2}{2} $.

Step 3: Combine the results of the integration into a single expression: $ \frac{1}{8}t^4 - \frac{1}{4}t^3 + \frac{1}{5}t^2 $. This represents the antiderivative of the given function.

Step 4: Evaluate the definite integral by substituting the limits of integration $ t = 0 $ and $ t = 1 $ into the antiderivative. First, substitute $ t = 1 $: $ \frac{1}{8}(1)^4 - \frac{1}{4}(1)^3 + \frac{1}{5}(1)^2 $. Then, substitute $ t = 0 $: $ \frac{1}{8}(0)^4 - \frac{1}{4}(0)^3 + \frac{1}{5}(0)^2 $.

Step 5: Subtract the value of the antiderivative at $ t = 0 $ from the value at $ t = 1 $. This will yield the final result of the definite integral. Simplify the terms to find the numerical value.