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Lemma
Integration with the Power Rule Quadratics Difficulty 4

TMUA practice: Integration Power Rule, problem 5

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

The question
Suppose a<ba < b are real numbers. What is the maximum value of   ∫ab(34−x−x2)dx  \;\displaystyle\int_{a}^{b} \left(\tfrac{3}{4} - x - x^{2}\right) dx\; over all such pairs (a,b)(a, b)?
A 34\tfrac{3}{4}
B 43\tfrac{4}{3}
C 32\tfrac{3}{2}
D 23\tfrac{2}{3}

The correct answer is highlighted. B

Worked solution

Where is the integrand positive? Solve   34−x−x2>0\;\dfrac{3}{4} - x - x^{2} > 0, i.e.   x2+x−34<0\;x^{2} + x - \dfrac{3}{4} < 0. By the quadratic formula, the roots are   x=−1±1+32=−1±22\;x = \dfrac{-1 \pm \sqrt{1 + 3}}{2} = \dfrac{-1 \pm 2}{2}, i.e.   x=12  \;x = \dfrac{1}{2}\; or   x=−32\;x = -\dfrac{3}{2}. The integrand is positive on   (−32, 12)\;(-\tfrac{3}{2},\, \tfrac{1}{2}).

Maximise. To maximise the integral, take   [a,b]=[−32, 12]  \;[a, b] = [-\tfrac{3}{2},\, \tfrac{1}{2}]\; (the full positive region). Extending beyond either endpoint would subtract a positive contribution (since the integrand becomes negative).

Compute. Antiderivative:   F(x)=3x4−x22−x33\;F(x) = \dfrac{3 x}{4} - \dfrac{x^{2}}{2} - \dfrac{x^{3}}{3}.

At x=12x = \tfrac{1}{2}:   F=3/24−1/42−1/83=38−18−124=624−124=524\;F = \dfrac{3/2}{4} - \dfrac{1/4}{2} - \dfrac{1/8}{3} = \dfrac{3}{8} - \dfrac{1}{8} - \dfrac{1}{24} = \dfrac{6}{24} - \dfrac{1}{24} = \dfrac{5}{24}.

At x=−32x = -\tfrac{3}{2}:   F=−98−9/42−−27/83=−98−98+98=−98=−2724\;F = -\dfrac{9}{8} - \dfrac{9/4}{2} - \dfrac{-27/8}{3} = -\dfrac{9}{8} - \dfrac{9}{8} + \dfrac{9}{8} = -\dfrac{9}{8} = -\dfrac{27}{24}.

Max integral:   F(1/2)−F(−3/2)=524−(−2724)=3224=43\;F(1/2) - F(-3/2) = \dfrac{5}{24} - \left(-\dfrac{27}{24}\right) = \dfrac{32}{24} = \dfrac{4}{3}.

This is option B.

The lesson: to maximise ∫abg(x) dx\int_{a}^{b} g(x)\, dx over all real a<ba < b, choose [a,b][a, b] to be the interval where g≥0g \ge 0.