#### To determine

**To show:**

2∫abfxf'xdx=fb2-fa2

#### Answer

2∫abfxf'xdx=fb2-fa2

#### Explanation

**Concept:**

i) The substitution rule: If u=g(x) is a differentiable function whose range is I and f is continuous on I, then ∫f(gx)g'xdx=∫f(u)du. Here g(x) is substituted as u and then differentiation g’(x)dx =du

ii)

∫xmdx=xm+1m+1+C, m is any real number

**Calculation:**

The given integral is

2∫abfxf'xdx=fb2-fa2

Consider the left side of the equation

2∫abfxf'xdx

Substitute fx=u therefore f'xdx=du,

And the limits of the integral becomes for x=a, u=fa for x=b, u=f(b)

Hence, the given integral becomes

2∫abfxf'xdx=2∫fafbudu

Now integrate

=2u22fafb

=(u2)fafb

Substitute the limits

=[fb]2-[fa]2

**Conclusion:**

Therefore, it is proved that 2∫abfxf'xdx=fb2-fa2