Question
Find the area of the region IN THE FIRST QUADRANT (upper right quadrant) bounded by the curves y=sin(x)cos(x)^2, y=2xcos(x^2) and y=4-4x.
You get:
a.)1.8467
b.) 0.16165
c.) 0.36974
d.) 1.7281
e.) 0.37859
Based on my calculations, I would say that the answer is e.) 0.37859. I am checking my answer.
You get:
a.)1.8467
b.) 0.16165
c.) 0.36974
d.) 1.7281
e.) 0.37859
Based on my calculations, I would say that the answer is e.) 0.37859. I am checking my answer.
Answers
That's the correct answer.
If you integrated between the end-points 0, 0.69275, 0.92811, you should have got the areas 0.28024 and 0.09835 respectively, which add up to 0.37859.
If you integrated between the end-points 0, 0.69275, 0.92811, you should have got the areas 0.28024 and 0.09835 respectively, which add up to 0.37859.
Thank you, I was just really unsure of my answer!
You're welcome! :)
Related Questions
Use a graphing calculator to graph f(x)=x^4-6x^3+11x^2-6x. Then use upper sums to approximate the a...
Find the area of the region IN THE FIRST QUADRANT (upper right quadrant) bounded by the curves y=sin...
Write the integral in one variable to find the volume of the solid obtained by rotating the firstâqu...
The region R is the region in the first quadrant bounded by the curves y = x^2-4x+4, x=0, and x=2....