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 6
Find the definite integral
∫20xx2+1
Incorrect
Loaded: 0%
Progress: 0%
0:00
First, form a fraction to balance the equation.
xx2+1 |
= |
x2x |
Differentiate x2+1 |
|
|
= |
12 |
xx=1 |
Use 12 as a constant to balance the integral.
|
|
12∫20f′(x)f(x)dx |
|
|
= |
12∫202xx2+1dx |
Substitute known values |
|
|
= |
[12ln(x2+1)]20 |
Finally, get the difference of the upper and lower limits substituted to the integral as x.
|
|
[12ln(x2+1)]20 |
|
|
= |
12[ln((2)2+1)-ln((0)2+1)] |
Substitute the limits |
|
|
= |
12[ln5-ln1] |
Evaluate |
|
|
= |
12ln5 |
-
Question 2 of 6
Find the definite integral
∫71x2x3+2
Round your answer to two decimal places
Incorrect
Loaded: 0%
Progress: 0%
0:00
First, form a fraction to balance the equation.
x2x3+2 |
= |
x23x2 |
Differentiate x3+2 |
|
|
= |
13 |
x2x2=1 |
Use 13 as a constant to balance the integral.
|
|
13∫71f′(x)f(x)dx |
|
|
= |
13∫713x2x3+2dx |
Substitute known values |
|
|
= |
[13ln(x3+2)]71 |
Finally, get the difference of the upper and lower limits substituted to the integral as x.
|
|
[13ln(x3+2)]71 |
|
|
= |
13[ln((7)3+2)-ln((1)3+2)] |
Substitute the limits |
|
|
= |
13[ln347-ln3] |
Evaluate |
|
|
= |
13[4.7507] |
Compute using calculator |
|
|
= |
1.58 |
Rounded to two decimal places |
-
Question 3 of 6
Find the definite integral
∫3112x−1
Incorrect
Loaded: 0%
Progress: 0%
0:00
First, form a fraction to balance the equation.
12x−1 |
= |
12 |
Differentiate 2x-1 |
Use 12 as a constant to balance the integral.
|
|
12∫31f′(x)f(x)dx |
|
|
= |
12∫3122x−1dx |
Substitute known values |
|
|
= |
[12ln(2x-1)]31 |
Finally, get the difference of the upper and lower limits substituted to the integral as x.
|
|
[12ln(2x-1)]31 |
|
|
= |
12[ln(2(3)-1)-ln(2(1)-1)] |
Substitute the limits |
|
|
= |
12[ln5-0] |
Evaluate |
|
|
= |
12ln5 |
-
Question 4 of 6
Find the area under the curve y=4x
Round your answer to three decimal places
Incorrect
Loaded: 0%
Progress: 0%
0:00
First, form a fraction to balance the equation.
4x |
= |
44 |
Differentiate x3+2 |
|
|
= |
4 |
41=4 |
We can see the limits indicated in the graph
Upper Limit |
= |
3 |
Lower Limit |
= |
1 |
Use 4 as a constant to balance the integral.
|
|
4∫31f′(x)f(x)dx |
|
|
= |
4∫711xdx |
Substitute known values |
|
|
= |
4[lnx]31 |
Finally, get the difference of the upper and lower limits substituted to the integral as x.
|
|
4[lnx]31 |
|
= |
4[ln(3)-ln(1)] |
Substitute the limits |
|
= |
4[ln3-ln1] |
Evaluate |
|
= |
4ln3 |
|
= |
4.394 |
Compute using calculator |
-
Question 5 of 6
Find the area under the curve y=xx2+1 going to the x-axis
Round your answer to two decimal places
Incorrect
Loaded: 0%
Progress: 0%
0:00
First, form a fraction to balance the equation.
xx2+1 |
= |
x2x |
Differentiate x3+2 |
|
|
= |
12 |
xx=1 |
We can see the limits indicated in the graph
Upper Limit |
= |
2 |
Lower Limit |
= |
0 |
Use 12 as a constant to balance the integral.
|
|
12∫20f′(x)f(x)dx |
|
|
= |
12∫202xx2+1dx |
Substitute known values |
|
|
= |
12[ln(x2+1)]20 |
Finally, get the difference of the upper and lower limits substituted to the integral as x.
|
|
12[ln(x2+1)]20 |
|
|
= |
12[ln((2)2+1)-ln((0)2+1)] |
Substitute the limits |
|
|
= |
12[ln5-ln1] |
Evaluate |
|
|
= |
12ln5 |
|
|
= |
0.80 |
Compute using calculator |
-
Question 6 of 6
Find the area under the curve y=lnx going to the x-axis
Incorrect
Loaded: 0%
Progress: 0%
0:00
Since we cannot integrate lnx, we can find the value of x and integrate it.
y |
= |
lnx |
ey |
= |
elnx |
Insert e as base to both sides |
ey |
= |
x |
elnx=x |
x |
= |
ey |
Therefore, the formula that we can use to find the area is ∫ eydy
We also need to find the limits for y
We can see on the graph that the upper limit on the x-axis is 3
Substitute this value to the equation to find the upper limit on the y-axis
y |
= |
lnx |
|
y |
= |
ln3 |
x=3 |
Upper Limit |
= |
ln3 |
Next, integrate and get the difference of the upper and lower limits substituted to the
integral as x.
|
|
∫ln30eydy |
|
= |
[ey]ln30 |
Integrate |
|
= |
e(ln3)-e(0) |
Substitute the limits |
|
= |
3-1 |
Anything raised to 0 is 1 |
|
= |
2 |
The area of the curve going to the y-axis is 2 units2
To find the area of the curve going to the x-axis, find the area of the whole rectangle and subtract the area of the curve going to the y-axis.
Area |
= |
length×width |
|
= |
ln3×3 |
|
= |
3ln3- 2 |