Reuse two basic integrals and a radical substitution

(a)(i) Find ∫2−x4x−x2−1dx\int\frac{2-x}{\sqrt{4x-x^2-1}}dx. marks

(ii) Find ∫14x−x2−1dx\int\frac1{\sqrt{4x-x^2-1}}dx. marks

(iii) Using (i) and (ii), find ∫x4x−x2−1dx\int\frac x{\sqrt{4x-x^2-1}}dx. marks

(b)(i) By using u=xu=\sqrt x, find ∫x1+xdx\int\frac{\sqrt x}{1+\sqrt x}dx. marks

(ii) Find the exact value of ∫09ln⁡(1+x)dx\int_0^9\ln(1+\sqrt x)dx, leaving the answer aln⁡b−c/ba\ln b-c/b, where a,b,ca,b,c are integers to be determined. marks

[2016/VJC/Promo/11]

Total marks: 14

Solution

If this part uses an earlier result, revisit (a)(i) Reverse chain rule or open all working together.

(a)(ii)

Completing the square

Find ∫14x−x2−1 dx\int\dfrac1{\sqrt{4x-x^2-1}}\,dx.

Explore this idea

Hint

  1. Complete the square: 4x−x2−1=3−(x−2)24x-x^2-1=3-(x-2)^2.

Answer · (a)(ii)

arcsin⁡((x−2)/3)+C\arcsin((x-2)/\sqrt3)+C.

Explore the idea

Open interactive exploration

A U-shaped curve with vertical asymptotes

Drag x. The shading shows the selected interval; the lower curves compare the numerical signed integral with the change in the antiderivative.

Upper limit xx = 2.83
Reset the controls

How the shaded areas add up

The limits run left to right. Area above the axis adds; area below the axis subtracts.

Above-axis contribution
0.496
Below-axis contribution
0
Signed integral
0.496

All three bars share one scale. Positive contributions extend right of zero; negative contributions extend left.

Signed integral from 2.00 to x
0.496
sin⁻¹((x−2)/√3) − F(2)
0.496
agrees at displayed precision
Formula − integral
0
0.00% of the integral’s magnitude
Compare the running integral on a graph
  • Integrand
  • Numerical signed integral from the start to x
  • Part of the integrand above the axis
  • Part of the integrand below the axis
  • F(x) − F(start), from the antiderivative
Key idea

The whole area under the curve from end to end is π\pi: the total change of sin⁡−1\sin^{-1} from −1-1 to 11.

Ideas to try
The area between the asymptotes is exactly π\pi.