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# Anti derivatives

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### Anti derivatives

1. 1. Michelle OwsleyTahniyat Farooqi Cyra Byramji
2. 2. Derivative Anti-derivativessin(x) -cos(x)cos(x) sin(x)sec2(x) tan(x)csc2(x) -cot(x)sec(x)tan(x) sec(x)csc(x)cot(x) -csc(x) Xn+1 Xn n+1 f’(g(x))g’(x) f(g(x)) f’(x) is also the same as the integral symbol ∫
3. 3.  Derive: f(x)=3x2+2x Answer: f’(x)= 6x+2 Now Anti-Derive: ∫6x+2dx Answer: f(x)=3x2+2x+c Where did the “c” come from?  The “c” means any constant. When you derive a constant in a function, the derivative is 0,so when anti-deriving always add “+c” at the end because you cannot assume whether or not there was a constant in the original function, and by adding “c” you are making sure you didn’t leave any numbers out of the function. RULES
4. 4.  Anti-derive the following:  ∫sin(x)dx   -cos(x) + c  ∫csc2(X)dx   -cot(x) + c  ∫sec(x)tan(x)dx   sec(x) + c  ∫4x + cos(x)dx   2x2 + sin(x) + c  Is there a rule for “4x” (one something to a power)? YES! If you don’t remember click the button  4x= 4x2 = 2x2 2 RULES
5. 5.  ∫3x2 - 7x + 4 - 5sec2(x)dx  Step by Step If you forgot your rules…  Rule for 3x2? Yes  x3  Rule for 7x? Yes  7x2 2  Rule for 4? Yes  4x  Rule for 5sec2(x)? Yes  5tan(x)  Put it all together…  X3 - 7x2 + 4x -5tan(x) + c 2 RULES
6. 6.  If the function u=g(x) has a continuous derivative on the closed interval [a,b] and f is continuous on the range of g, then  ∫ab f(g(x))g’(x)dx = ∫g(a)g(b) f(u)du Anti-derive:  ∫(x2+4) 9 (2x)dx  Is there are rule for this one? Of course, it’s the Product Rule…WRONG! If this is what you were thinking then Click!  You have to use u-substitution to solve this problem. Derivative  u= x2+4 and du=2xdx  Now Substitute!
7. 7.  u= x2+4 du=2xdx ∫(x2+4) 9 (2x)dx ∫(u) 9du  This is your new equation, now is there a rule for this? Yes, so Anti-derive. u 10 + c Not finished yet…now plug back in your original numbers. 10 (x2+4)10 + c  This is the final answer 10 RULES
8. 8.  ∫5cos(5x)dx  u=5x  du=5dx  ∫cos(u)du  sin(u) + c  sin(5x) + c ∫x(5x 2 - 3)7dx  u=5x 2 – 3  du=10xdx but there is not 10x in the problem?!?  That’s ok, 10 is a coefficient so just move it! 1/10du=xdx  Now that we have everything…rewrite the problem  ∫u7 (1/10)du  (1/10)u8  u8 + c 8 80  Don’t forget to plug the original numbers back in …  (5x 2 - 3)8 + c 80 RULES