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Find the area of the region by given curves y=3sinx,y=e^5x,x=0 and x=π/2

+2 votes

Please help!!

asked Dec 25, 2012 in CALCULUS by homeworkhelp Mentor

3 Answers

+3 votes

Sketch region enclosed by y = 3sinx, y = e^(5x), and x = 0 , x =π/2

The two curves intersect at point (0,π/2)

and from x = 0 to x = π/2

Integrating from x = 0 to x = π/2, we get

A = ∫(0 to π/2) 3sinx- e^(5x)) dx

A = (- 3cosx - e^(5x)/5) |(0 to π/2)

A = (- 3cos(π/2)- e^(5π/2)/5  +3cos0 - e^(5*0)/5)

A = (- 3(0)- e^(5π/2)/5  +3(1) - e^0/5)

A = (- e^(5π/2)/5  +3 - 1/5)

A = (- e^(5π/2)/5  +3 - 1/5)

answered Dec 27, 2012 by ashokavf Scholar
how do you find the points of intersection?

There are no points of intersection for given curves.

Graph the two curves and vertical lines :

Observe the graph

No points of intersection exists between x = 0 and x = π/2.

Hence we find the area of the region between two curves and the vertical lines x = 0 and x = π/2.

0 votes

 

Sketch region enclosed by y = 3sinx, y = e^(5x), and x = 0 , x =π/2

The two curves intersect at point (0,π/2)

and from x = 0 to x = π/2

Integrating from x = 0 to x = π/2, we get

A = ∫(0 to π/2) 3sinx- e^(5x)) dx

A = (- 3cosx - e^(5x)/5) |(0 to π/2)

A = (- 3cos(π/2)e^(5π/2)/5  +3cos0 + e^(5*0)/5)

A = (- 3(0)e^(5π/2)/5  +3(1) + e^0/5)

A = (e^(5π/2)/5  +3 + 1/5)

A = (e^(5π/2)/5  +3 + 1/5)

A = (- e^(5pi/2)/5  +16/5)

 

answered May 13, 2013 by jeevitha Novice

This is the correct solution.

0 votes

Area of Region between two curves :

If f(x) and g(x) are continuous on [a, b] and g(x) ≤ f(x) for all x in [a, b], then the area of the region bounded by the graphs of f(x) and g(x) vertical lines x = a and x = b is A = ∫(a to b) [ f(x) - g(x) ] dx.

The two curves are f(x) = 3 sin(x), g(x) = e5x, x = 0 and x = π/2.

If x = 0 then f(x) = 3 sin(0) = 0 and g(x) = e5(0) = 0.

If x = π/2 then f(x) = 3 sin(π/2) = 3 and g(x) = e5(π/2).

Since f(x) ≤ g(x) for all x in [0, π/2], then A = ∫(a to b) [ g(x) - f(x) ] dx.

A = ∫(0 to π/2) [ e5x - 3 sin(x) ] dx

A =  e5 [x2/2] (0 to π/2) - 3 [cos(x)] (0 to π/2)

A =  e5 [π2/8] - 3 [1]

A =  (e5π2 - 24)/8.

answered Jun 23, 2014 by casacop Expert

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