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