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

asked Dec 25, 2012 in CALCULUS

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)

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.

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)

This is the correct solution.

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.