Sunday

March 29, 2015

March 29, 2015

Posted by **Rebecca** on Thursday, February 9, 2012 at 8:48pm.

- Calculus 2 -
**Steve**, Thursday, February 9, 2012 at 9:58pmsince e^6x is greater than e^2x on [0,1], I'd integrate with respect to x.

Int(e^6x - e^2x dx)[0,1]

= 1/6 e^6x - 1/2 e^2x [0,1]

= (1/6 e^6 - 1/2 e^2) - (1/6 - 1/2)

= 1/6 e^2 (e^4 - 3) + 1/3

= 63.8

**Answer this Question**

**Related Questions**

calculus - Sketch the region enclosed by x+y^2=42 and x+y=0. Decide whether to ...

Calculus - Sketch the region enclosed by the curves x= 49-y^2 and x = y^2 - 49. ...

calculus - Sketch the region enclosed by 2y=5sqrtx, y=5 and 2y+3x=8. Decide ...

Calculus - Sketch the region enclosed by the given curves. Decide whether to ...

Calculus - Sketch the region enclosed by the given curves. Decide whether to ...

Calculus - Sketch the region enclosed by the given curves. Decide whether to ...

CALCULUS:) - Sketch the region enclosed by the given curves. Decide whether to ...

Calculus - sketch the region enclosed by the given curves. Decide whether to ...

calculus - Sketch the region enclosed by the given curves. Decide whether to ...

calculus - Sketch the region enclosed by the given curves. Decide whether to ...