- #1

- 32

- 0

∫ (v^2 - e^(3v)) dv.

I did

∫(V^2-e^(3v)) dv

∫(v^2)dv - I (e^(3v) )dv

∫(v^3)/3- (e^(3v))/3

∫(v^3-e^(3v))/3

Did I so it right?

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- Thread starter rowdy3
- Start date

- #1

- 32

- 0

∫ (v^2 - e^(3v)) dv.

I did

∫(V^2-e^(3v)) dv

∫(v^2)dv - I (e^(3v) )dv

∫(v^3)/3- (e^(3v))/3

∫(v^3-e^(3v))/3

Did I so it right?

- #2

Mark44

Mentor

- 35,028

- 6,774

∫ (v^2 - e^(3v)) dv.

I did

∫(V^2-e^(3v)) dv

∫(v^2)dv - I (e^(3v) )dv

∫(v^3)/3- (e^(3v))/3

∫(v^3-e^(3v))/3

Did I so it right?

Not quite. After you antidifferentiate the integral sign should be gone. Also, you need to include the constant of integration and you should use = to connect expressions that are equal.

∫(v^2-e^(3v)) dv

= ∫(v^2)dv - ∫ (e^(3v) )dv

= (v^3)/3- (e^(3v))/3 + C

This could also be written as

(1/3)v

or as (1/3)(v

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