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| FTC #1
| | <math>y=\int_{1-3x}^{1}\frac{u^3}{1+u^2} du</math> |
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| <math>G(x)=f^\prime(x)</math> or in other words <math>\frac{d}{dx}[\int\limits_{a(x)}^{b(x)}F(x)dx]</math> is <math>\ b^\prime(x)*f(b(x))-a^\prime(x)*f(a(x))</math>
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| <math>y=\int\limits_{1-3x}^{1}\frac{u^3}{(1+u^2)} du</math> | | <math> |
| | \frac{d}{dx}(y)=\frac{d}{dx}\left(\int_{1-3x}^{1}\frac{u^3}{1+u^2}\,du\right) = (0)\cdot\frac{(1)^3}{1+(1)^2} |
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| so
| | -(-3)\cdot\frac{(1-3x)^3}{1+(1-3x)^2} |
| <math>y=\int\limits_{1-3x}^{1}\frac{1}{(1+u^2)}u^3, du</math>
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| using the formula we get y=<math>(0)*f(1)-(-3)*f(1-3x)</math>
| | </math> |
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| which is equal to <math>(3)*f(1-3x)</math>
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| which is=<math>3*(1-3x)^3*\frac{1}{(1+(1-3x)^2)}</math>
| | <math> |
| or simplified to <math>\frac{3*(1-3x)^3}{(1+(1-3x)^2)}</math>
| | \text{Therefore, } y' = \frac{3(1-3x)^3}{1+(1-3x)^2} |
| | | </math> |
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| FTC #2 (not done yet)
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| <math>\int\limits_{a}^{b}f(x)dx</math> is equal to <math>F(b)-F(a)</math> Where F is the antiderivative of f such that <math>F^\prime=f</math>
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| <math>y=\int\limits_{1-3x}^{1}\frac{x^3}{(1+u^2)} dx</math>
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| <math>y=\int\limits_{1-3x}^{1}\frac{x^3}{(1+u^2)} dx</math>
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Latest revision as of 20:31, 6 September 2022
