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Lemma
Integration with the Power Rule Difficulty 4

TMUA practice: Integration Power Rule, problem 3

A TMUA-calibre integration power rule problem at difficulty 4 of 5, with the full worked solution.

The question
Which of the following is equal to   ∫(2−3x)4 dx \;\displaystyle\int (2 - 3 x)^{4}\, dx\,?
A (2−3x)515+C\dfrac{(2 - 3 x)^{5}}{15} + C
B −(2−3x)515+C-\dfrac{(2 - 3 x)^{5}}{15} + C
C (2−3x)55+C\dfrac{(2 - 3 x)^{5}}{5} + C
D −3(2−3x)3+C-3 (2 - 3 x)^{3} + C
E −(2−3x)55+C-\dfrac{(2 - 3 x)^{5}}{5} + C

The correct answer is highlighted. B

Worked solution

Use the linear shift u=2−3xu = 2 - 3 x, so du=−3 dxdu = -3\, dx and dx=−du/3dx = -du/3. (Note the negative sign from the negative coefficient.)

∫(2−3x)4 dx  =  ∫u4⋅−du3  =  −13⋅u55+C  =  −(2−3x)515+C.\int (2 - 3 x)^{4}\, dx \;=\; \int u^{4} \cdot \frac{-du}{3} \;=\; -\frac{1}{3} \cdot \frac{u^{5}}{5} + C \;=\; -\frac{(2 - 3 x)^{5}}{15} + C.

This is option B.

Cross-check by differentiation. ddx[−(2−3x)515]=−5(2−3x)4⋅(−3)15=15(2−3x)415=(2−3x)4\dfrac{d}{dx}\left[-\dfrac{(2 - 3 x)^{5}}{15}\right] = -\dfrac{5 (2 - 3 x)^{4} \cdot (-3)}{15} = \dfrac{15 (2 - 3 x)^{4}}{15} = (2 - 3 x)^{4} ✓.

Why the other options fail.

  • A, (2−3x)5/15+C(2 - 3 x)^{5}/15 + C: sign error — forgot the minus from dx=−du/3dx = -du/3.
  • C, (2−3x)5/5+C(2 - 3 x)^{5}/5 + C: forgot the factor 1/31/3 from 1/a=1/(−3)1/a = 1/(-3), kept the magnitude.
  • D, −3(2−3x)3+C-3 (2 - 3 x)^{3} + C: differentiated the original instead of integrating.
  • E, −(2−3x)5/5+C-(2 - 3 x)^{5}/5 + C: kept the sign but lost the factor 1/31/3.

The lesson: ∫(ax+b)n dx=(ax+b)n+1/(a(n+1))\int (a x + b)^{n}\, dx = (a x + b)^{n+1}/(a (n+1)). When a<0a < 0, the negative propagates through.