Which of the following integrals represents the area of the region bounded by the curves and between and ?
Table of contents
- 0. Functions7h 52m
- Introduction to Functions16m
- Piecewise Functions10m
- Properties of Functions9m
- Common Functions1h 8m
- Transformations5m
- Combining Functions27m
- Exponent rules32m
- Exponential Functions28m
- Logarithmic Functions24m
- Properties of Logarithms34m
- Exponential & Logarithmic Equations35m
- Introduction to Trigonometric Functions38m
- Graphs of Trigonometric Functions44m
- Trigonometric Identities47m
- Inverse Trigonometric Functions48m
- 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
- 10. Physics Applications of Integrals 3h 16m
- 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
- 16. Parametric Equations & Polar Coordinates7h 58m
8. Definite Integrals
Introduction to Definite Integrals
Struggling with Calculus?
Join thousands of students who trust us to help them ace their exams!Watch the first videoMultiple Choice
What is the average (mean) value of the function over the interval ?
A
B
C
D

1
Step 1: Recall the formula for the average value of a function f(t) over an interval [a, b]: \( \text{Average Value} = \frac{1}{b-a} \int_{a}^{b} f(t) \, dt \). Here, \( f(t) = 3t^3 - t^2 \), \( a = 1 \), and \( b = 2 \).
Step 2: Set up the integral \( \int_{1}^{2} (3t^3 - t^2) \, dt \). This involves finding the antiderivative of \( 3t^3 - t^2 \).
Step 3: Compute the antiderivative of \( 3t^3 - t^2 \). The antiderivative of \( 3t^3 \) is \( \frac{3t^4}{4} \), and the antiderivative of \( -t^2 \) is \( -\frac{t^3}{3} \). Combine these to get \( \frac{3t^4}{4} - \frac{t^3}{3} \).
Step 4: Evaluate the definite integral \( \int_{1}^{2} (3t^3 - t^2) \, dt \) by substituting the limits of integration into the antiderivative. This means calculating \( \left[ \frac{3t^4}{4} - \frac{t^3}{3} \right]_{1}^{2} \), which involves substituting \( t = 2 \) and \( t = 1 \) and subtracting.
Step 5: Divide the result of the definite integral by \( b-a \), which is \( 2-1 = 1 \), to find the average value of the function over the interval. The final average value is \( \frac{1}{1} \int_{1}^{2} (3t^3 - t^2) \, dt \).
Watch next
Master Definition of the Definite Integral with a bite sized video explanation from Patrick
Start learningRelated Videos
Related Practice
Multiple Choice
21
views
Introduction to Definite Integrals practice set
