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

(a)(i)

Reverse chain rule

Find ∫2−x4x−x2−1 dx\int\dfrac{2-x}{\sqrt{4x-x^2-1}}\,dx.

Explore this idea

Hint

  1. The derivative of 4x−x2−14x-x^2-1 is 4−2x=2(2−x)4-2x=2(2-x).

Answer · (a)(i)

4x−x2−1+C\sqrt{4x-x^2-1}+C.

Explore the idea

A radical with domain 2−√3 < x < 2+√3

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
Below-axis contribution
-0.209
Signed integral
-0.209

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

Signed integral from 2.00 to x
-0.209
√(4x−x²−1) − F(2)
-0.209
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 antiderivative is the radical itself, symmetric about x=2x=2.

Ideas to try
The integrand is an odd function about x=2x=2. What does that say about the area?
(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

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.
(a)(iii)

Linear combination

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

Explore this idea

Hint

  1. Write x=−(2−x)+2x=-(2-x)+2.

Answer · (a)(iii)

−4x−x2−1+2arcsin⁡((x−2)/3)+C-\sqrt{4x-x^2-1}+2\arcsin((x-2)/\sqrt3)+C.

All a-integrands: 2−√3<x<2+√3.

Explore the idea

Combining the two basic integrals

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
1.202
Below-axis contribution
0
Signed integral
1.202

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

Signed integral from 2.00 to x
1.202
antiderivative − F(2)
1.202
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 new integrand is a combination, so its antiderivative is the same combination of the two earlier answers.

Ideas to try
Compare with the previous two labs: what is the sum of them with the factors?
(b)(i)

Radical substitution

By using u=xu=\sqrt x, find ∫x1+x dx\int\dfrac{\sqrt x}{1+\sqrt x}\,dx.

Explore this idea

Hint

  1. x=u2x=u^2 so Differentiate x=u2x=u^2 to obtain dx=2u dudx=2u\,du. Equivalently, for x>0x>0, differentiate u=xu=\sqrt x and invert its derivative. Use u≥0u\ge0; then 1+u>01+u>0, which justifies the logarithm without modulus.

  2. Divide to get a polynomial plus a simple fraction.

Answer · (b)(i)

x−2x+2ln⁡(1+x)+Cx-2\sqrt x+2\ln(1+\sqrt x)+C.

x≥0x\ge0; derivative on x>0.

Explore the idea

A curve rising towards 1

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 = 4.50
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
2.534
Below-axis contribution
0
Signed integral
2.534

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

Signed integral from 0.00 to x
2.534
antiderivative − F(0)
2.534
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

For large xx the area grows almost linearly, because the integrand tends to 1.

Ideas to try
The integrand approaches 1 for large xx: what does that make the area?
(b)(ii)

Integration by parts with a given integral

Find the exact value of ∫09ln⁡(1+x) dx\int_0^9\ln(1+\sqrt x)\,dx, in the form aln⁡b−cba\ln b-\dfrac cb.

Explore this idea

Hint

  1. Parts with u=ln⁡(1+x)u=\ln(1+\sqrt x) and dv=dxdv=dx.

  2. The remaining integral is part (b)(i).

Answer · (b)(ii)

16ln⁡2−3/216\ln2-3/2; a=16,b=2,c=3a=16,b=2,c=3.

Explore the idea

The area from 0 to 9

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 = 9.00
Try an example

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
9.59
Below-axis contribution
0
Signed integral
9.59

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

Signed integral from 0.00 to x
9.59
antiderivative − F(0)
9.59
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 exact area is 16ln⁡2−3216\ln2-\frac32.

Ideas to try
Drag xx to 9: the readout is 16ln⁡2−32≈9.5916\ln2-\frac32\approx9.59.