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CBSE Sample Paper 2021-22 (Term 2) · Section C · 4 marks

Q11: ∫−12∣x3−3x2+2x∣ dx\int_{-1}^{2} |x^3 - 3x^2 + 2x|\,dx

Sample 2021-22 T2Ch 7 · Integrals
The question

Evaluate: ∫−12∣x3−3x2+2x∣ dx\displaystyle\int_{-1}^{2} |x^3 - 3x^2 + 2x|\,dx

Exam answer
Let I=∫−12∣x3−3x2+2x∣ dx\text{Let } I = \int_{-1}^{2} |x^3 - 3x^2 + 2x|\,dx

Here x3−3x2+2x=x(x−1)(x−2)x^3 - 3x^2 + 2x = x(x - 1)(x - 2)⁠. It is ≤0\le 0 on [−1,0][-1, 0]⁠, ≥0\ge 0 on [0,1][0, 1] and ≤0\le 0 on [1,2][1, 2]⁠.

Why?

To remove a modulus, first find where the expression inside changes sign. Take out xx⁠, then factorise the quadratic:

x3−3x2+2x=x(x2−3x+2)=x(x−1)(x−2)\begin{aligned} &x^3 - 3x^2 + 2x \\ &= x(x^2 - 3x + 2) \\ &= x(x - 1)(x - 2) \end{aligned}

For the quadratic, find two numbers with product 2 and sum −3-3⁠: they are −1-1 and −2-2⁠.

The zeros are 0,1,20, 1, 2⁠. Test one point in each piece: at x=−12x = -\frac12 the value is −158-\frac{15}{8}⁠, at x=12x = \frac12 it is 38\frac38⁠, and at x=32x = \frac32 it is −38-\frac38⁠.

x=−12: (−12)(−32)(−52)=−158x=12: (12)(−12)(−32)=38x=32: (32)(12)(−12)=−38\begin{aligned} x = -\tfrac12&:\ \left(-\tfrac12\right)\left(-\tfrac32\right)\left(-\tfrac52\right) \\ &\quad = -\tfrac{15}{8} \\ x = \tfrac12&:\ \left(\tfrac12\right)\left(-\tfrac12\right)\left(-\tfrac32\right) \\ &\quad = \tfrac38 \\ x = \tfrac32&:\ \left(\tfrac32\right)\left(\tfrac12\right)\left(-\tfrac12\right) \\ &\quad = -\tfrac38 \end{aligned}

The three brackets are xx⁠, x−1x - 1 and x−2x - 2 at that point.

I=∫−10∣x(x−1)(x−2)∣ dx+∫01∣x(x−1)(x−2)∣ dx+∫12∣x(x−1)(x−2)∣ dx\begin{aligned} I &= \int_{-1}^{0} |x(x - 1)(x - 2)|\,dx \\ &\quad + \int_0^1 |x(x - 1)(x - 2)|\,dx \\ &\quad + \int_1^2 |x(x - 1)(x - 2)|\,dx \end{aligned}
Why?

Split the interval at the zeros 00 and 11 that lie inside [−1,2][-1, 2]⁠, using

∫abf dx=∫acf dx+∫cbf dx\int_a^b f\,dx = \int_a^c f\,dx + \int_c^b f\,dx
=−∫−10(x3−3x2+2x) dx+∫01(x3−3x2+2x) dx−∫12(x3−3x2+2x) dx\begin{aligned} &= -\int_{-1}^{0} (x^3 - 3x^2 + 2x)\,dx \\ &\quad + \int_0^1 (x^3 - 3x^2 + 2x)\,dx \\ &\quad - \int_1^2 (x^3 - 3x^2 + 2x)\,dx \end{aligned}
Why?

Where the expression is negative, ∣f(x)∣=−f(x)|f(x)| = -f(x)⁠. Where it is positive, ∣f(x)∣=f(x)|f(x)| = f(x)⁠.

Watch out: leaving out a minus sign makes the pieces partly cancel instead of adding up.

=−[x44−x3+x2]−10+[x44−x3+x2]01−[x44−x3+x2]12\begin{aligned} &= -\left[\frac{x^4}{4} - x^3 + x^2\right]_{-1}^{0} \\ &\quad + \left[\frac{x^4}{4} - x^3 + x^2\right]_0^1 \\ &\quad - \left[\frac{x^4}{4} - x^3 + x^2\right]_1^2 \end{aligned}
Why?

Integrate term by term with ∫xn dx=xn+1n+1\int x^n\,dx = \frac{x^{n + 1}}{n + 1}⁠:

∫(x3−3x2+2x) dx=x44−x3+x2+C\begin{aligned} &\int (x^3 - 3x^2 + 2x)\,dx \\ &= \frac{x^4}{4} - x^3 + x^2 + C \end{aligned}

Term by term:

∫x3 dx=x44∫3x2 dx=3⋅x33=x3∫2x dx=2⋅x22=x2\begin{aligned} \int x^3\,dx &= \frac{x^4}{4} \\ \int 3x^2\,dx &= 3 \cdot \frac{x^3}{3} = x^3 \\ \int 2x\,dx &= 2 \cdot \frac{x^2}{2} = x^2 \end{aligned}
=−(0−94)+(14−0)−(0−14)=94+14+14\begin{aligned} &= -\left(0 - \frac94\right) + \left(\frac14 - 0\right) \\ &\quad - \left(0 - \frac14\right) \\ &= \frac94 + \frac14 + \frac14 \end{aligned}
Why?

With F(x)=x44−x3+x2F(x) = \frac{x^4}{4} - x^3 + x^2⁠:

F(−1)=14+1+1=94F(0)=0F(1)=14−1+1=14F(2)=4−8+4=0\begin{aligned} F(-1) &= \tfrac14 + 1 + 1 = \tfrac94 \\ F(0) &= 0 \\ F(1) &= \tfrac14 - 1 + 1 = \tfrac14 \\ F(2) &= 4 - 8 + 4 = 0 \end{aligned}

With the substitutions written out:

F(−1)=(−1)44−(−1)3+(−1)2=14+1+1=94F(1)=(1)44−(1)3+(1)2=14−1+1=14F(2)=(2)44−(2)3+(2)2=164−8+4=4−8+4=0\begin{aligned} F(-1) &= \frac{(-1)^4}{4} - (-1)^3 + (-1)^2 \\ &= \frac14 + 1 + 1 = \frac94 \\[4pt] F(1) &= \frac{(1)^4}{4} - (1)^3 + (1)^2 \\ &= \frac14 - 1 + 1 = \frac14 \\[4pt] F(2) &= \frac{(2)^4}{4} - (2)^3 + (2)^2 \\ &= \frac{16}{4} - 8 + 4 \\ &= 4 - 8 + 4 = 0 \end{aligned}

Each bracket is F(upper)−F(lower)F(\text{upper}) - F(\text{lower})⁠.

−(F(0)−F(−1))=−(0−94)=94F(1)−F(0)=14−0=14−(F(2)−F(1))=−(0−14)=14\begin{aligned} -\left(F(0) - F(-1)\right) &= -\left(0 - \frac94\right) = \frac94 \\ F(1) - F(0) &= \frac14 - 0 = \frac14 \\ -\left(F(2) - F(1)\right) &= -\left(0 - \frac14\right) = \frac14 \end{aligned}

Add them: 94+14+14=9+1+14=114\frac94 + \frac14 + \frac14 = \frac{9 + 1 + 1}{4} = \frac{11}{4}⁠.

∴ I=114\therefore\ I = \dfrac{11}{4}

Where this comes from

Chapter 7, Integrals: Definite integral of a modulus function.

Question type: A modulus inside the integral

How to spot it: click to see

Type 6. A modulus inside the integral: split at the corner

Spot it: bars inside the integral, like ∣x+3∣|x + 3| or x∣x∣x|x|⁠.

Here: Factorise the inside: x(x−1)(x−2)x(x - 1)(x - 2) is 0 at 0, 1 and 2. The zeros 0 and 1 lie inside [−1,2][-1, 2]⁠, so split there and put a minus sign on the pieces where it is negative.

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