## anonymous 5 years ago calculate the integral of the following using:

1. anonymous

the standard form $\int\limits_{}^{}f'(x)e ^{f(x)}dx=e ^{f(x)}+c , or \int\limits_{}^{}e ^{u}(du/dx)dx = e ^{u}+c$

2. anonymous

$(a) 2xe ^{x ^{2}+3}$ $(b)(10x-2)e ^{5x ^{2}-2x}$ $(c)(3x+2)e ^{3x ^{2}+4x+1}$ $(d)(x ^{2}-2x)e ^{x ^{3}-3x ^{2}}$

3. watchmath

What have you try with these problems?

4. anonymous

yea i have i can do (a),(b) but c and d i got the wrong answers

5. watchmath

You need to adjust a little bit for c) and d). For example in c) if you use $$u=3x^2+4x$$ then $$du=6x+4=2(3x+2)\, dx$$. So your integral become $$\frac{1}{2}\int e^u\, du$$

6. anonymous

oh i understand thanks but what happens to the (3x+2) inside the bracket

7. watchmath

$$(3x+2)\,dx$$ become $$\frac{1}{2}du$$

8. anonymous

ok so how about this one $(6x ^{2}-8x+6)e ^{x ^{3}-2x ^{2}+3x-5}$

9. watchmath

what is the derivative of $$x^3-2x^2+3x-5$$ ?

10. anonymous

3x^2-4x+3

11. watchmath

Good! How that derivative relates to $$6x^2-8x+6$$ ?

12. anonymous

its half of it right

13. anonymous

the coefficients are doubled

14. watchmath

Yes great! Then for $$f(x)=x^3-2x^2+3x-5$$ the integral is $$\frac{1}{2}\int f'(x)e^{f(x)}\,dx$$

15. watchmath

16. watchmath

So $$2\int f'(x)e^{f(x)}\,dx$$.

17. anonymous

oh i see how it works out now lol:) ok thanks ill have shot at the others thankyou

18. watchmath

Great! :D