calculate the integral of the following using:

Question

anonymous · Answer

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

anonymous · Answer

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

watchmath · Answer

What have you try with these problems?

anonymous · Answer

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

watchmath · Answer

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 $\frac{1}{2}\int e^u\, du$

anonymous · Answer

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

watchmath · Answer

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

anonymous · Answer

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

watchmath · Answer

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

anonymous · Answer

3x^2-4x+3

watchmath · Answer

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

anonymous · Answer

its half of it right

anonymous · Answer

the coefficients are doubled

watchmath · Answer

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

watchmath · Answer

Sorry your were right :)

watchmath · Answer

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

anonymous · Answer

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

watchmath · Answer

Great! :D