Evaluating integral (definite)

Hi all

Is my solution correct?

Evaluate "integration sign" [(2e^x)-SQRT(x)] between limits b=10 and a=1.

Integrating each tem separately:

**2e^x.dx,**

re-arranging: e^x.2.dx................(1)

let u=x, du=1.dx, 2du=2dx

substituting u=x and 2du=2dx back into (1) gives

(e^u). 2du

Integrated gives:

(e^u). 2

2e^x as u=x.

Integrating **SQRT(x)**

SQRT(x)= x^(1/2)

Integrated gives:

(x^3/2)/3/2

Simplyfying

[(2x^(3/2))/3]

Combining both integrals

[2e^x]-[(2x^(3/2))/3] and evaluating between limits of b=10 and a=1

{[2e^x]-[(2x^(3/2))/3]}-{[2e^x]-[(2x^(3/2))/3]}

{[2e^10]-[(2(10)^(3/2))/3]}-{[2e^(1)]-[(2(1)^(3/2))/3]}

{44,052-21.1}-{5.4-0.66)

44,026?

thanks

John

Re: Evaluating integral (definite)

Quote:

Originally Posted by

**celtic1234** Hi all

Is my solution correct?

Evaluate "integration sign" [(2e^x)-SQRT(x)] between limits b=10 and a=1.

Integrating each tem separately:

**2e^x.dx,**

re-arranging: e^x.2.dx................(1)

let u=x, du=1.dx, 2du=2dx

substituting u=x and 2du=2dx back into (1) gives

(e^u). 2du

Integrated gives:

(e^u). 2

2e^x as u=x.

Integrating **SQRT(x)**

SQRT(x)= x^(1/2)

Integrated gives:

(x^3/2)/3/2

Simplyfying

[(2x^(3/2))/3]

Combining both integrals

[2e^x]-[(2x^(3/2))/3] and evaluating between limits of b=10 and a=1

{[2e^x]-[(2x^(3/2))/3]}-{[2e^x]-[(2x^(3/2))/3]}

{[2e^10]-[(2(10)^(3/2))/3]}-{[2e^(1)]-[(2(1)^(3/2))/3]}

{44,052-21.1}-{5.4-0.66)

44,026?

thanks

John

Why would you bother doing a u substitution for $\displaystyle \displaystyle \int{2e^x\,dx}$? Surely you know that $\displaystyle \displaystyle \int{k\,f(x)\,dx} = k\int{f(x)\,dx}$ for constant $\displaystyle \displaystyle k$...

It looks ok otherwise :)