Use the supplied derivative, then split the sign

(a) Write down ddx(tan⁡(x2))\frac d{dx}(\tan(x^2)). Hence find ∫x3sec⁡2(x2)dx\int x^3\sec^2(x^2)dx. marks

(b) Evaluate ∫04x2∣x−3∣dx\int_0^4x^2|x-3|dx without the use of the graphic calculator. marks

[2010/DHS/Prelim/I/6 (part)]

Total marks: 7

Solution

(a) derivative

Chain rule

Write down ddx(tan⁡x2)\dfrac d{dx}\left(\tan x^2\right).

Hint

  1. Use ddxtan⁡u=sec⁡2u⋅u′\frac d{dx}\tan u=\sec^2u\cdot u' with u=x2u=x^2.

Answer · (a) derivative

2xsec⁡2(x2)2x\sec^2(x^2).

(a) integral

Integration by parts using a given derivative

Hence find ∫x3sec⁡2(x2) dx\int x^3\sec^2(x^2)\,dx.

Explore this idea

Hint

  1. Write x3sec⁡2(x2)=12x2⋅2xsec⁡2(x2)x^3\sec^2(x^2)=\frac12x^2\cdot2x\sec^2(x^2) so that the supplied derivative appears.

  2. Let u=x2u=x^2 and dvdx=2xsec⁡2(x2)\frac{dv}{dx}=2x\sec^2(x^2), so v=tan⁡(x2)v=\tan(x^2).

Answer · (a) integral

12[x2tan⁡(x2)−ln⁡∣sec⁡(x2)∣]+C\frac12[x^2\tan(x^2)-\ln|\sec(x^2)|]+C.

On intervals with cos(x²)≠0.

Explore the idea

Before the first pole of sec²(x²)

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 = 0.60
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
0.035
Below-axis contribution
0
Signed integral
0.035

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

Signed integral from 0.00 to x
0.035
antiderivative − F(0)
0.035
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 integrand is odd, so its signed left and right integrals cancel on a symmetric interval. The running integral from 0 is even: it has the same value at opposite probe points within one interval free of poles.

Ideas to try
The integrand blows up at x2=π2x^2=\frac\pi2, about ±1.25\pm1.25. Slide close to it.
(b)

Integrating an absolute value

Evaluate ∫04x2∣x−3∣ dx\int_0^4x^2|x-3|\,dx without a calculator.

Explore this idea

Hint

  1. ∣x−3∣=3−x|x-3|=3-x for x<3x<3 and x−3x-3 for x≥3x\ge3.

  2. Split the integral at x=3x=3.

Answer · (b)

27/227/2.

Explore the idea

The area splits at x = 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 = 4.00
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
13.5
Below-axis contribution
0
Signed integral
13.5

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

Signed integral from 0.00 to x
13.5
piecewise antiderivative
13.5
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

Both pieces count as positive area: 274+274=272\frac{27}4+\frac{27}4=\frac{27}2.

Ideas to try
Drag xx past 3: the antiderivative changes formula there.