Posted: August 26th, 2021
Student’s Name
Instructor’s Name
Course
Date
Calculus
Question 1
Limited by the function 𝑦 = −2𝑥^2, the x-axis and the points x= -1 and x=2 (10 points)
Solutions
The area under the curve between the two points is attained by using definite integral. The issue can be solved using integration concepts in calculus (Courant, 133). Given the following;
This can be divided into two parts:
=-2/3+-16/3 =-18/3
=-6 units sq. The area is negative because the curve is below the x-axis.
Question 2
Limited by the function 𝑦 = 6𝑥^2𝑒^x^3, the x-axis and the points x=0 and x=2 (20 points)
Solutions
The solution is ascertained using integration by substitution method as follows;
Let u=x^2
du=2xdx
Using limits
=327.589 units sq.
(Fernandez, 107)
Question 3
The limited by the function of the x-axis and the points x=1 and x=3.
(Tip: You will need a scientific calculator for the last step) (20 points)
Solutions
Applying the same concept as in question 2, then;
Let u =x^3-3x
Introducing limits
= -2.022 units sq.
Question 4
Limited by the function 𝑦 = cos 2 , the x-axis and the points x=15 and x=30, where 15 and 30 are degrees. (Tip: you will need a scientific calculator for finding some trigonometric values.
(Set up: Sexagesimal Degrees) (25 points)
Solutions
Integral by substitution concept is applied as follows;
Let u=2x
Introducing limits (Fernandez, 112);
= 0.1830 units sq.
Question 5
Limited by the function 𝑦 = 𝑥𝑐𝑜𝑠𝑥, the x-axis and the points x=30 and x=60 where 30 and 60 are degrees (25 points)
Solutions
( Masani, 94)
Will use integration by parts formula
Let u=x =du=dx
Introducing limits
= 36.5955 Units Sq.
Works Cited
Courant, Richard. Differential and Integral Calculus. New York: John Wiley & Sons, 2011. Print.
Fernandez, Oscar E. Calculus simplified. Princeton, New Jersey: Princeton University Press, 2019. Print.
Masani, Pesi R., R. C. Patel, and D. J. Patil. Elementary calculus. New York: Academic Press, 2009. Print.
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