To determine

To evaluate:

The indefinite integral ∫x1-x2 dx

Answer

-13(1-x2)32+C

Explanation

1) Concept:

i) The substitution rule

If u=g(x) is a differentiable function whose range is an interval I and f is continuous on I, then ∫f(gx)g'xdx=∫f(u)du.

ii) Indefinite integral

∫xn dx=xn+1n+1+C   (n≠-1)

2) Given:

∫x1-x2 dx

3) Calculation:

Here, use the substitution method because the differential of the function 1-x2 is -2xdx

Substitute u=1-x2

Differentiate u=1-x2 with respect to x

du=-2xdx

As x dx is a part of the integration, solving for x dx by dividing both side by 2.

-du2=xdx

By using concept i),

substitute u=1-x2,  xdx=-du2 .∫x1-x2 dx=∫u-du2

By rearranging

=-12∫udu

By using concept ii),

=-12u12+112+1+C

=-12u3232+C

Cancelling out common factor

=-u323+C

Rearranging

=-u323+C

Resubstitute u=1-x2

=-13(1-x2)32+C

Conclusion:

Therefore,

∫x1-x2 dx=-13(1-x2)32+C