To determine

To find:

The derivative of function gx=∫2x3xu2-1u2+1 du

Answer

g'x=-24x2-14x2+1+39x2-19x2+1

Explanation

1) Concept:

i) Properties of definite integral:

∫ba fxdx=-∫ab fxdx

2) Fundamental theorem of Calculus, Part 1:

If f is continuous on a, b, then the function g defined by

gx=∫0x ft dt   a≤x≤b

gx is continuous on a,b, and g'x=f(x)

3) Given:

gx=∫2x3xu2-1u2+1 du

4) Calculation:

Rewrite given integral as (by using hint given),

∫2x3xu2-1u2+1 du=∫2x0u2-1u2+1 du+∫03xu2-1u2+1 du …(1)

Now by using concept i) (Property of definite integral),

=-∫02xu2-1u2+1 du+∫03xu2-1u2+1 du

gx=-∫02xu2-1u2+1 du+∫03xu2-1u2+1 du

Differentiating with respect to x,

g'x=ddx-∫02xu2-1u2+1 du+ddx∫03xu2-1u2+1 du

By using Fundamental theorem of Calculus, Part 1,

g'x=-u2-1u2+1ddtu- u2-1u2+1ddt0+u2-1u2+1ddtt- u2-1u2+1ddt0

Substitute the values of limit (u=2x and 0 in first integral,u=3x and 0 in second integral),

g'x=-2x2-12x2+1ddx2x- 0-10+1ddx0+3x2-13x2+1ddx3x- 0-10+1ddx0

=-2x2-12x2+1ddx2x+0+3x2-13x2+1ddx3x- 0

=-22x2-12x2+1+33x2-13x2+1

g'x=-24x2-14x2+1+39x2-19x2+1

Conclusion:

Therefore, the derivative of given integral is -24x2-14x2+1+39x2-19x2+1