Fundamental Theorem of Calculus, Part 1 practice problems
Differentiate an integral whose upper limit is a function of x. 70 problems with worked solutions; in 70 of them every
equation is proved by a computer algebra system.
20 worked problems
Try each one before opening the solution. Every line is labeled with what checked it. How we verify
✓ Nihil obstat Lines: 1 proved, 2 reviewed. The answer was also checked a second way, without looking at the solution. Reviewers found nothing wrong with the explanation. The full receipt ·
How we verify
✓ Nihil obstat Lines: 1 proved, 2 reviewed. The answer was also checked a second way, without looking at the solution. Reviewers found nothing wrong with the explanation. The full receipt ·
How we verify
By FTC Part 1 and the chain rule, d/dx ∫ from a to u(x) of g(t) dt = g(u(x)) u'(x).
Reviewed
\[ \frac{d}{d x} x^{3} = 3 x^{2} \]
u'(x).✓ Proved
\[ \frac{3 x^{2}}{x^{12} + 1} \]
g(u(x)) u'(x).Reviewed
Answer \( \frac{3 x^{2}}{x^{12} + 1} \)
✓ Nihil obstat Lines: 1 proved, 2 reviewed. The answer was also checked a second way, without looking at the solution. Reviewers found nothing wrong with the explanation. The full receipt ·
How we verify
By FTC Part 1 and the chain rule, d/dx ∫ from a to u(x) of g(t) dt = g(u(x)) u'(x).
Reviewed
\[ \frac{d}{d x} x = 1 \]
u'(x).✓ Proved
\[ 1 \frac{1}{x^{4} + 1} = \frac{1}{x^{4} + 1} \]
g(u(x)) u'(x).Reviewed
Answer \( \frac{1}{x^{4} + 1} \)
✓ Nihil obstat Lines: 1 proved, 2 reviewed. The answer was also checked a second way, without looking at the solution. Reviewers found nothing wrong with the explanation. The full receipt ·
How we verify
By FTC Part 1 and the chain rule, d/dx ∫ from a to u(x) of g(t) dt = g(u(x)) u'(x).
Reviewed
\[ \frac{d}{d x} x^{3} = 3 x^{2} \]
u'(x).✓ Proved
\[ 3 x^{2} \ln{\left(x^{6} + 2 \right)} \]
g(u(x)) u'(x).Reviewed
Answer \( 3 x^{2} \ln{\left(x^{6} + 2 \right)} \)
✓ Nihil obstat Lines: 1 proved, 2 reviewed. The answer was also checked a second way, without looking at the solution. Reviewers found nothing wrong with the explanation. The full receipt ·
How we verify
By FTC Part 1 and the chain rule, d/dx ∫ from a to u(x) of g(t) dt = g(u(x)) u'(x).
Reviewed
\[ \frac{d}{d x} x = 1 \]
u'(x).✓ Proved
\[ \sqrt{x^{2} + 2} \]
g(u(x)) u'(x).Reviewed
Answer \( \sqrt{x^{2} + 2} \)
✓ Nihil obstat Lines: 1 proved, 2 reviewed. The answer was also checked a second way, without looking at the solution. Reviewers found nothing wrong with the explanation. The full receipt ·
How we verify
✓ Nihil obstat Lines: 1 proved, 2 reviewed. The answer was also checked a second way, without looking at the solution. Reviewers found nothing wrong with the explanation. The full receipt ·
How we verify
By FTC Part 1 and the chain rule, d/dx ∫ from a to u(x) of g(t) dt = g(u(x)) u'(x).
Reviewed
\[ \frac{d}{d x} x = 1 \]
u'(x).✓ Proved
\[ \ln{\left(x^{2} + 2 \right)} \]
g(u(x)) u'(x).Reviewed
Answer \( \ln{\left(x^{2} + 2 \right)} \)
✓ Nihil obstat Lines: 1 proved, 2 reviewed. The answer was also checked a second way, without looking at the solution. Reviewers found nothing wrong with the explanation. The full receipt ·
How we verify
By FTC Part 1 and the chain rule, d/dx ∫ from a to u(x) of g(t) dt = g(u(x)) u'(x).
Reviewed
\[ \frac{d}{d x} x = 1 \]
u'(x).✓ Proved
\[ \sin{\left(x^{2} \right)} \]
g(u(x)) u'(x).Reviewed
Answer \( \sin{\left(x^{2} \right)} \)
✓ Nihil obstat Lines: 1 proved, 2 reviewed. The answer was also checked a second way, without looking at the solution. Reviewers found nothing wrong with the explanation. The full receipt ·
How we verify
By FTC Part 1 and the chain rule, d/dx ∫ from a to u(x) of g(t) dt = g(u(x)) u'(x).
Reviewed
\[ \frac{d}{d x} x = 1 \]
u'(x).✓ Proved
\[ e^{- x^{2}} \]
g(u(x)) u'(x).Reviewed
Answer \( e^{- x^{2}} \)
✓ Nihil obstat Lines: 1 proved, 2 reviewed. The answer was also checked a second way, without looking at the solution. Reviewers found nothing wrong with the explanation. The full receipt ·
How we verify
✓ Nihil obstat Lines: 1 proved, 2 reviewed. The answer was also checked a second way, without looking at the solution. Reviewers found nothing wrong with the explanation. The full receipt ·
How we verify
By FTC Part 1 and the chain rule, d/dx ∫ from a to u(x) of g(t) dt = g(u(x)) u'(x).
Reviewed
\[ \frac{d}{d x} x^{3} = 3 x^{2} \]
u'(x).✓ Proved
\[ 3 x^{2} e^{- x^{6}} \]
g(u(x)) u'(x).Reviewed
Answer \( 3 x^{2} e^{- x^{6}} \)
✓ Nihil obstat Lines: 1 proved, 2 reviewed. The answer was also checked a second way, without looking at the solution. Reviewers found nothing wrong with the explanation. The full receipt ·
How we verify
By FTC Part 1 and the chain rule, d/dx ∫ from a to u(x) of g(t) dt = g(u(x)) u'(x).
Reviewed
\[ \frac{d}{d x} 3 x = 3 \]
u'(x).✓ Proved
\[ 3 \sin{\left(9 x^{2} \right)} \]
g(u(x)) u'(x).Reviewed
Answer \( 3 \sin{\left(9 x^{2} \right)} \)
✓ Nihil obstat Lines: 1 proved, 2 reviewed. The answer was also checked a second way, without looking at the solution. Reviewers found nothing wrong with the explanation. The full receipt ·
How we verify
By FTC Part 1 and the chain rule, d/dx ∫ from a to u(x) of g(t) dt = g(u(x)) u'(x).
Reviewed
\[ \frac{d}{d x} 3 x = 3 \]
u'(x).✓ Proved
\[ 81 x^{3} \cos{\left(3 x \right)} \]
g(u(x)) u'(x).Reviewed
Answer \( 81 x^{3} \cos{\left(3 x \right)} \)
✓ Nihil obstat Lines: 1 proved, 2 reviewed. The answer was also checked a second way, without looking at the solution. Reviewers found nothing wrong with the explanation. The full receipt ·
How we verify
✓ Nihil obstat Lines: 1 proved, 2 reviewed. The answer was also checked a second way, without looking at the solution. Reviewers found nothing wrong with the explanation. The full receipt ·
How we verify
By FTC Part 1 and the chain rule, d/dx ∫ from a to u(x) of g(t) dt = g(u(x)) u'(x).
Reviewed
\[ \frac{d}{d x} 3 x = 3 \]
u'(x).✓ Proved
\[ \frac{3}{81 x^{4} + 1} \]
g(u(x)) u'(x).Reviewed
Answer \( \frac{3}{81 x^{4} + 1} \)
✓ Nihil obstat Lines: 1 proved, 2 reviewed. The answer was also checked a second way, without looking at the solution. Reviewers found nothing wrong with the explanation. The full receipt ·
How we verify
By FTC Part 1 and the chain rule, d/dx ∫ from a to u(x) of g(t) dt = g(u(x)) u'(x).
Reviewed
\[ \frac{d}{d x} x^{3} = 3 x^{2} \]
u'(x).✓ Proved
\[ 3 x^{2} \sqrt{x^{6} + 1} \]
g(u(x)) u'(x).Reviewed
Answer \( 3 x^{2} \sqrt{x^{6} + 1} \)
✓ Nihil obstat Lines: 1 proved, 2 reviewed. The answer was also checked a second way, without looking at the solution. Reviewers found nothing wrong with the explanation. The full receipt ·
How we verify
By FTC Part 1 and the chain rule, d/dx ∫ from a to u(x) of g(t) dt = g(u(x)) u'(x).
Reviewed
\[ \frac{d}{d x} x^{3} = 3 x^{2} \]
u'(x).✓ Proved
\[ 3 x^{2} \sin{\left(x^{6} \right)} \]
g(u(x)) u'(x).Reviewed
Answer \( 3 x^{2} \sin{\left(x^{6} \right)} \)
✓ Nihil obstat Lines: 1 proved, 2 reviewed. The answer was also checked a second way, without looking at the solution. Reviewers found nothing wrong with the explanation. The full receipt ·
How we verify
By FTC Part 1 and the chain rule, d/dx ∫ from a to u(x) of g(t) dt = g(u(x)) u'(x).
Reviewed
\[ \frac{d}{d x} x = 1 \]
u'(x).✓ Proved
\[ x^{3} \cos{\left(x \right)} \]
g(u(x)) u'(x).Reviewed
Answer \( x^{3} \cos{\left(x \right)} \)
✓ Nihil obstat Lines: 1 proved, 2 reviewed. The answer was also checked a second way, without looking at the solution. Reviewers found nothing wrong with the explanation. The full receipt ·
How we verify
By FTC Part 1 and the chain rule, d/dx ∫ from a to u(x) of g(t) dt = g(u(x)) u'(x).
Reviewed
\[ \frac{d}{d x} x^{2} = 2 x \]
u'(x).✓ Proved
\[ 2 x^{7} \cos{\left(x^{2} \right)} \]
g(u(x)) u'(x).Reviewed
Answer \( 2 x^{7} \cos{\left(x^{2} \right)} \)
✓ Nihil obstat Lines: 1 proved, 2 reviewed. The answer was also checked a second way, without looking at the solution. Reviewers found nothing wrong with the explanation. The full receipt ·
How we verify
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