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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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| [[5.3 The Fundamental Theorem of Calculus/1|1]]
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| [[5.3 The Fundamental Theorem of Calculus/3|3]]
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| [[5.3 The Fundamental Theorem of Calculus/5|5]]
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| [[5.3 The Fundamental Theorem of Calculus/7|7]]
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| [[5.3 The Fundamental Theorem of Calculus/8|8]]
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| [[5.3 The Fundamental Theorem of Calculus/9|9]]
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| [[5.3 The Fundamental Theorem of Calculus/10|10]]
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| [[5.3 The Fundamental Theorem of Calculus/11|11]]
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| [[5.3 The Fundamental Theorem of Calculus/13|13]]
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| [[5.3 The Fundamental Theorem of Calculus/15|15]]
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| [[5.3 The Fundamental Theorem of Calculus/17|17]]
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| [[5.3 The Fundamental Theorem of Calculus/19|19]]
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| [[5.3 The Fundamental Theorem of Calculus/20|20]]
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| [[5.3 The Fundamental Theorem of Calculus/21|21]]
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| [[5.3 The Fundamental Theorem of Calculus/23|23]]
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| [[5.3 The Fundamental Theorem of Calculus/25|25]]
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| [[5.3 The Fundamental Theorem of Calculus/27|27]]
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| [[5.3 The Fundamental Theorem of Calculus/28|28]]
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| [[5.3 The Fundamental Theorem of Calculus/29|29]]
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| [[5.3 The Fundamental Theorem of Calculus/31|31]]
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| [[5.3 The Fundamental Theorem of Calculus/33|33]]
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| [[5.3 The Fundamental Theorem of Calculus/35|35]]
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| [[5.3 The Fundamental Theorem of Calculus/37|37]]
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| [[5.3 The Fundamental Theorem of Calculus/39|39]]
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| [[5.3 The Fundamental Theorem of Calculus/41|41]]
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| [[5.3 The Fundamental Theorem of Calculus/53|53]]
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Failed to parse (Conversion error. Server ("https://en.wikipedia.org/api/rest_") reported: "Cannot get mml. Server problem."): {\displaystyle y=\int _{1-3x}^{1}{\frac {u^{3}}{1+u^{2}}}du}
Failed to parse (MathML with SVG or PNG fallback (recommended for modern browsers and accessibility tools): Invalid response ("Math extension cannot connect to Restbase.") from server "https://en.wikipedia.org/api/rest_v1/":): {\displaystyle \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} -(-3)\cdot\frac{(1-3x)^3}{1+(1-3x)^2} }
Failed to parse (MathML with SVG or PNG fallback (recommended for modern browsers and accessibility tools): Invalid response ("Math extension cannot connect to Restbase.") from server "https://en.wikipedia.org/api/rest_v1/":): {\displaystyle \text{Therefore, } y' = \frac{3(1-3x)^3}{1+(1-3x)^2} }