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2007-03-23 21:08:00 · 6 answers · asked by Obaid K 1 in Science & Mathematics Mathematics

6 answers

∫ ( x^3 e^(5x) dx )

This involves using integration by parts twice.

Let u = x^3. dv = e^(5x) dx
du = 3x^2 dx. v = (1/5)e^(5x)

(1/5)x^3 e^(5x) - ∫ ( 3x^2 (1/5) e^(5x) dx )

Factor out the constants from the integral.

(1/5)x^3 e^(5x) - (3/5) ∫ (x^2 e^(5x) dx )

Use parts again.

Let u = x^2. dv = e^(5x) dx
du = 2x dx. v = (1/5)e^(5x)

(1/5)x^3 e^(5x) - (3/5) [ (1/5)x^2 e^(5x) - ∫ (2x (1/5) e^(5x) dx )]

Distribute the (-3/5).

(1/5)x^3 e^(5x) - (3/25)x^2 e^(5x) + (3/5) ∫ 2x (1/5) e^(5x) dx )

Pull out the constants from the integral.

(1/5)x^3 e^(5x) - (3/25) x^2 e^(5x) + (3/5)(2/5)∫ (x e^(5x) dx)

(1/5)x^3 e^(5x) - (3/25) x^2 e^(5x) + (6/25) ∫(x e^(5x) dx)

Use parts (last time! I promise!)

Let u = x. dv = e^(5x) dx.
du = dx. v = (1/5)e^(5x)

(1/5)x^3 e^(5x) - (3/25) x^2 e^(5x) + (6/25) [(1/5)x e^(5x) -
∫ ( (1/5)e^(5x) ) ]

Distribute the 6/25.

(1/5)x^3 e^(5x) - (3/25) x^2 e^(5x) + (6/125) x e^(5x) -
(6/125) ∫ ( e^(5x) dx )

Now the integral is trivial.

(1/5)x^3 e^(5x) - (3/25) x^2 e^(5x) + (6/125) x e^(5x) -
(6/125) (1/5) e^(5x) + C

(1/5)x^3 e^(5x) - (3/25) x^2 e^(5x) + (6/125) x e^(5x) -
(6/625) e^(5x) + C

2007-03-23 21:22:46 · answer #1 · answered by Puggy 7 · 0 1

f(x) = x^3*e^5x
h(x) = x^3
g(x) = e^5x => g'(x) = 5 g(x) dx
f(x) = h(x) * g(x)
= [h(x) g'(x)/5] dx
using integration by parts :
i will use | | for integration symbol

|| f(x)dx = || [h(x) g'(x)/5] dx = 1/5[h(x)g(x) - || h'(x)g(x)dx]
= 1/5[h(x)g(x) - ||3x^2 * g(x)dx]
Let k(x) = 3x^2
Then :: ||3x^2 * g(x)dx = || [k(x) g'(x)/5] dx =
1/5[k(x)g(x) - ||6x * g(x)dx]

Let r(x) = 6x
Then :: ||6x * g(x)dx = || [r(x) g'(x)/5] dx =
1/5[r(x)g(x) - ||6* g(x)dx]

And we know that ||6* g(x)dx = 6/5 e^5x

|| f(x)dx
= 1/5[h(x)g(x) - 1/5[k(x)g(x) - 1/5[r(x)g(x) - 6/5 e^5x]]]
= 1/5 h(x)g(x) -1/25k(x)g(x) +1/125 r(x)g(x) - 6/625 e^5x
= (x^3 e^5x)/5 - (3x^2 e^5x)/25 + (6x e^5x)/125 - 6* e^5x /625
= e^5x/625 [ 125 x^3 - 75 x^2 + 30 x -6]

Answer is [e^5x * (125 x^3 - 75 x^2 + 30 x -6)]/625

2007-03-24 04:51:44 · answer #2 · answered by Nishit V 3 · 0 0

=x^3e^5x /5 - integrate 3/5 x^2e^5x
= " - 3/25 x^2 e^5x +3/5 integrate2/5x e^5x
= " - " + 6/125 xe^5x -6/25 integrate e^5x
=1/5x^3 e^5x - 3/25 x^2e^5x +6/125 xe^5x-6/125 e^5x

2007-03-24 04:22:53 · answer #3 · answered by SS 2 · 0 0

Hint: Integrate by parts

2007-03-24 04:23:54 · answer #4 · answered by Demiurge42 7 · 0 0

You will need to do integration by parts three times. Each time you lower the x power by 1.

2007-03-24 04:19:46 · answer #5 · answered by mathsmanretired 7 · 0 1

(3x^2)(e^5x)+(5x^3)(e^5x)

2007-03-24 04:23:44 · answer #6 · answered by swt4life03 2 · 0 1

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