Information
You have already completed the quiz before. Hence you can not start it again.
You must sign in or sign up to start the quiz.
You have to finish following quiz, to start this quiz:
Loading...
-
Question 1 of 4
Find the exact area bounded by the curves y=x2 and y=3x
Incorrect
Loaded: 0%
Progress: 0%
0:00
Remember
By integrating the difference of two functions, the area can be obtained.
We are asked to find the area between the curves y=3x and y=x2
A |
= |
∫ (3x-x2) dx |
Subtract the lower curve from the higher curve |
Take the points of intersection as the limits
A |
= |
∫30(3x−x2)dx |
Points intersect at x=0 and x=3 |
Find the Indefinite Integral using the Power Rule
∫(3x−x2)dx |
= |
3(x1+11+1)−(x2+12+1) |
Apply the Power Rule |
|
|
= |
3(x22)–x33 |
Simplify |
|
|
= |
3x22–x33 |
Find the Definite Integral
A |
= |
∫30(3x−x2)dx |
|
|
= |
[3x22–x33]30 |
|
|
= |
[3(32)2-333]–[3(02)2–033] |
Substitute the upper (3) and lower limits (0) |
|
|
= |
[272-273]–[0-0] |
Simplify |
|
|
= |
272–9 |
|
|
= |
272–182 |
|
|
= |
92 |
-
Question 2 of 4
Find the exact area bounded by the curves y=x-1 and y=x2-6x+5
Incorrect
Loaded: 0%
Progress: 0%
0:00
Remember
By integrating the difference of two functions, the area can be obtained.
We are asked to find the area between the curves y=x-1 and y=x2-6x+5
A |
= |
∫ (x-1-(x2-6x+5)) dx |
Subtract the lower curve from the higher curve |
|
= |
∫ (-6+7x-x2) dx |
Take the points of intersection as the limits
A |
= |
∫61(−6+7x−x2)dx |
Points intersect at x=1 and x=6 |
Find the Indefinite Integral using the Power Rule
∫(−6+7x−x2)dx |
= |
−6(x0+10+1)+7(x1+11+1)–(x2+12+1) |
Apply the Power Rule |
|
|
= |
-6x11+7x22-x33 |
Simplify |
|
|
= |
-6x+7x22-x33 |
Find the Definite Integral
A |
= |
∫61(−6+7x−x2)dx |
|
|
= |
[−6x+7x22–x33]61 |
|
|
= |
[-6(6)+7(62)2-633]–[-6(1)+7(12)2–133] |
Substitute the upper (6) and lower limits (1) |
|
|
= |
[-36+7(36)2-2163]–[-6+72-13] |
Simplify |
|
|
= |
18-(-176) |
|
|
= |
1256 |
-
Question 3 of 4
Find the exact area bounded by the curves y=10-x2 and y=4-x
Incorrect
Loaded: 0%
Progress: 0%
0:00
Remember
By integrating the difference of two functions, the area can be obtained.
We are asked to find the area between the curves y=10-x2 and y=4-x
A |
= |
∫ (10-x2-(4-x)) dx |
Subtract the lower curve from the higher curve |
|
= |
∫ (-x2+x+6) dx |
Take the points of intersection as the limits
A |
= |
∫3−2(−x2+x+6)dx |
Points intersect at x=-2 and x=3 |
Find the Indefinite Integral using the Power Rule
∫(−x2+x+6)dx |
= |
−(x2+12+1)+(x1+11+1)+6(x0+10+1) |
Apply the Power Rule |
|
|
= |
-x33+x22+6x11 |
Simplify |
|
|
= |
-x33+x22+6x |
Find the Definite Integral
A |
= |
∫3−2(−x2+x+6)dx |
|
|
= |
[−x33+x22+6x]3−2 |
|
|
= |
[-333+322+6(3)]–[-(-2)33+(-2)22+6(-2)] |
Substitute the upper (3) and lower limits (-2) |
|
|
= |
[-9+92+18]–[83+2-12] |
Simplify |
|
|
= |
272–(-223) |
|
|
= |
1256 |
-
Question 4 of 4
Find the exact area bounded by the curves y=x2-6x+8 and y=-x2+4x-4
Incorrect
Loaded: 0%
Progress: 0%
0:00
Remember
By integrating the difference of two functions, the area can be obtained.
We are asked to find the area between the curves y=x2-6x+8 and y=-x2+4x-4
A |
= |
∫ (x2-6x+8-(-x2+4x-4)) dx |
Subtract the lower curve from the higher curve |
|
= |
∫ (2x2-10x+12) dx |
Take the points of intersection as the limits. We can divide the area into two.
A |
= |
A1+A2 |
|
= |
∫20 (2x2-10x+12) dx+∫32 (2x2-10x+12) dx |
Get individual integrals |
Find the Indefinite Integral using the Power Rule
∫(2x2−10x+12)dx |
= |
(2x2+12+1)−10(x1+11+1)+12(x0+10+1) |
Apply the Power Rule |
|
|
= |
2x33-10x22+12(x1)1 |
Simplify |
|
|
= |
2x33-5x2+12x |
Solve for A1 using Definite Integral
A1 |
= |
∫20(2x2−10x+12)dx |
|
|
= |
[2x33−5x2+12x]20 |
|
|
= |
[2(23)3–5(22)+12(2)]–[2(03)3–5(02)+12(0)] |
Substitute the upper (2) and lower limits (0) |
|
|
= |
[163-20+24]–[0] |
Simplify |
|
A1 |
= |
283square units |
A2 |
= |
∫32(2x2−10x+12)dx |
|
|
= |
[2x33−5x2+12x]32 |
|
|
= |
[2(33)3–5(32)+12(3)]–[2(23)3–5(22)+12(2)] |
Substitute the upper (3) and lower limits (2) |
|
|
= |
[543-45+36]–[163–20+24] |
Simplify |
|
|
= |
9–283 |
|
|
= |
-13 |
Get the absolute value |
|
A2 |
= |
13square units |
A |
= |
A1+A2 |
|
= |
283+13 |
Substitute calculated values |
|
A |
= |
293square units |