Answer to Question #264666 in Real Analysis for Nikhil rawat

Question

State the second mean value theorem of integrability. Verify it for the function f and g is defined by f(x)= 6x and g(x)= -5x on [3,4].

Answer 1

The second mean value theorem of integrability states that if the integral of a(x) and b(x) are continuous on [f,g] and b(x)$\ge$ 0 then,

$\int_f^ga(x)b(x)dx=a(f)\int_f^hb(x)dx+a(g)\int_h^gb(x)dx$

Given:

f(x)=a(x)=6x

g(x)=b(x)=-5x

Where [f,g]=[3,4]

$\int_3^4(6x)(-5x)dx=6(3)\int_3^h-5xdx+6(4)\int_h^4-5xdx$

$\implies \int_3^4-30x^2dx=18\int_3^h(-5)dx+24\int_h^4(-5x)dx$

$\implies -30(\frac{x^3}{3})_3^4=-90(\frac{x^2}{2})_3^h+-20(\frac{x^2}{2})_h^4$

$\implies -30(\frac{64-27}{3})=-90(\frac{h^2-9}{2})-120(\frac{16-h^2}{2})$

$\implies -370=-45(h^2-9)-60(16-h^2)$

$\implies -370=-45h^2+405-960+60h^2$

$\implies -370+960-405=15h^2$

185=15h²

h²=12.33

$\implies h=3.511$

as it defines the function