
Integration by Substitution - choosing a u
Authored by Susan Goodling
Mathematics
12th Grade
Used 69+ times

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16 questions
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1.
MULTIPLE CHOICE QUESTION
1 min • 1 pt
Identify the u and the du:
u=10x, du = 10dx
u = 5x2 + 1, du = 10xdx
u = (5x2 +1)2, du = 10xdx
u = (5x2 +1)2, du = 2(5x2 + 1) 10xdx
Answer explanation
The correct choice is u = 5x² + 1, du = 10xdx. When you substitute, it will become u to the 2nd power, and the x with cancel with the du.
2.
MULTIPLE CHOICE QUESTION
1 min • 1 pt
u = x2
u = x
u = x - 4
Answer explanation
The correct choice is u= x - 4. When you make the substitution, the x2 does not cancel, so you will need to solve for x in terms of u, then simplify.
3.
MULTIPLE CHOICE QUESTION
1 min • 1 pt
Identify the u and the du.
u = -x2, du = -2x dx
u = 1 - x2, du = -2xdx
u = -2x, du = -2dx
Answer explanation
The correct choice is u = 1 - x², du = -2xdx. This substitution simplifies integration, as it directly relates to the derivative of u with respect to x, making it easier to work with in calculus problems.
4.
MULTIPLE CHOICE QUESTION
1 min • 1 pt
Identify the u and the du.
u = 3x2, du = -6x dx
u = x3 + 1, du = 3x2 dx
Answer explanation
The correct choice is u = x^3 + 1, du = 3x^2 dx. Here, u is a simple polynomial, and its derivative, du, is calculated using the power rule, confirming the relationship between u and du.
5.
MULTIPLE CHOICE QUESTION
1 min • 1 pt
Identify the u and the du.
Answer explanation
The correct choice is u=(4+1/x^2) and du=-2/x^3 dx. This is derived from differentiating u with respect to x, applying the chain rule.
6.
MULTIPLE CHOICE QUESTION
1 min • 1 pt
Identify the u and the du.
u = 1+2x, du = 2dx
u = 2, du = 0dx
u = 2x, du = 2dx
Answer explanation
The correct choice is u = 1+2x, du = 2dx. Here, u is defined as a function of x, and du represents its derivative with respect to x, which is calculated as 2dx.
7.
MULTIPLE CHOICE QUESTION
1 min • 1 pt
u = x
Answer explanation
The correct choice is u = 2x^2, so that the du = 4x will cancel the x on the top. You will then create inverse sine as the integral to solve.
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