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IB Mathematics: Applications and Interpretation HL · 13.5: Integration · 13.5.2
Anti-differentiation: reversing the power rule Problem set B Answers
Skills
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- (a) \(f(x)=\frac{1}{10}x^{10}+c\)
- (b) \(f(x)=-\frac{1}{4}x^{-4}+c\)
- (c) \(f(x)=3x+c\)
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- (a) \(y=2x^{9} + c\)
- (b) \(y=\frac{1}{3}x^{7} + c\)
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- (a) \(y=2x^{-3} + c\)
- (b) \(y=-\frac{1}{3}x^{-5} + c\)
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- (a) \(y=x^{7} + 2x^{4} - 2x + c\)
- (b) \(y=\frac{2}{25}x^{5} - \frac{3}{4}x^{2} + c\)
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\(x^{3} + 8x - 2x^{-5} + c\)
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\(-\frac{3}{4}y^{-3} - \frac{1}{5}y^{-6} + c\)
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- (a) \(f(x)=5x^{-3} + c\)
- (b) \(f(x)=-3x^{-6} + c\)
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- (a) \(f(x)=8x^{3/2} + c\)
- (b) \(f(x)=12x^{1/2} + c\)
In context
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- (a) \(V=3t^{3} + 4t^{2} - 6t+c\)
- (b) c = 0
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\(B=3r^{2} + 5r^{-3}+c\)
Stretch
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\(f(x)=6x^{3/2} - 16x^{1/2} - \frac{9}{2}x^{-2}+c\)
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Priya differentiated instead of integrating. Correct: \(-\frac{9}{2}x^{-2}+c\)
jmaths.xyz · AI HL · 13.5.2 Anti-differentiation: reversing the power rule