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Worked Examples · Example 12

Q.Evaluate the following:

(a) ∫−11x2x3+1 dx\int_{-1}^{1} x^2\sqrt{x^3+1}\,dx
(b) ∫−30xx+4 dx\int_{-3}^{0} x\sqrt{x+4}\,dx
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Two definite integrals cracked by substitution with changed limits: (a) =429=\dfrac{4\sqrt{2}}{9},

(b) =−9415=-\dfrac{94}{15}.

Power rule ∫un du=un+1n+1+C\displaystyle\int u^{n}\,du=\frac{u^{n+1}}{n+1}+C (n≠−1n\neq-1). When substituting u=g(x)u=g(x), replace the xx-limits by the matching uu-limits.

(a) ∫−11x2x3+1 dx\displaystyle\int_{-1}^{1} x^2\sqrt{x^3+1}\,dx

  1. Put u=x3+1⇒du=3x2 dx⇒x2 dx=du3u=x^3+1\Rightarrow du=3x^2\,dx\Rightarrow x^2\,dx=\dfrac{du}{3}.
  2. New limits: x=−1⇒u=0x=-1\Rightarrow u=0; x=1⇒u=2x=1\Rightarrow u=2.
  3. I=13∫02u du=13⋅23[u3/2]02=29(23/2−0).I=\dfrac{1}{3}\displaystyle\int_{0}^{2}\sqrt{u}\,du=\dfrac{1}{3}\cdot\dfrac{2}{3}\Big[u^{3/2}\Big]_{0}^{2}=\dfrac{2}{9}\big(2^{3/2}-0\big).
  4. 23/2=222^{3/2}=2\sqrt{2}, so I=29⋅22=429≈0.629.I=\dfrac{2}{9}\cdot 2\sqrt{2}=\dfrac{4\sqrt{2}}{9}\approx 0.629.

(b) ∫−30xx+4 dx\displaystyle\int_{-3}^{0} x\sqrt{x+4}\,dx

  1. Put u=x+4⇒x=u−4, dx=duu=x+4\Rightarrow x=u-4,\ dx=du; limits x=−3⇒u=1x=-3\Rightarrow u=1, x=0⇒u=4x=0\Rightarrow u=4.
  2. I=∫14(u−4)u du=∫14(u3/2−4u1/2) du=[25u5/2−83u3/2]14.I=\displaystyle\int_{1}^{4}(u-4)\sqrt{u}\,du=\int_{1}^{4}\big(u^{3/2}-4u^{1/2}\big)\,du=\Big[\tfrac{2}{5}u^{5/2}-\tfrac{8}{3}u^{3/2}\Big]_{1}^{4}. …

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