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Let $f(x): \mathbb{R} \to \mathbb{R^+} $ be the function:

$$ f(x)=\sqrt{17 x^2+x^4}, $$ whose plot is:

plotted

By integrating the function, I got:
$$ F(x)=\frac{\left(x^2 \left(x^2+17\right)\right)^{3/2}}{3 x^3}+C, $$ whose plot is:
enter image description here


When I evaluate the definite integral:

$$ \int _{-1}^2(f x)d x=F(2)-F(-1), $$

I get $$ 18 \sqrt{2}+7 \sqrt{21}. $$

But I'm unsure: since I have the discontinuity, should I have splitted up the interval and calculated the integral from $-1$ to the limits $\to 0$ from left side. And $2$ to the limits $\to 0$ from the right side?

Dal
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Onizuka
  • 135

1 Answers1

1

Hints:

$$\int_{-1}^2\sqrt{17x^2+x^4}dx=\int_{-1}^2|x|\sqrt{17+x^2}dx=\int_{-1}^0-x\sqrt{17+x^2}dx+\int_{0}^2x\sqrt{17+x^2}dx=\int_{-1}^0-\frac12\sqrt{17+x^2}dx^2+\int_{0}^2\frac12\sqrt{17+x^2}dx^2.$$

Let $t=17+x^2$. Do you know what to do next step?

Paul
  • 20,553
  • I used the substitution method. A=1/3 (18^(3/2)-17^(3/2)) and B=1/3 (21^(3/2)-17^(3/2)). I=A+B = 1/3 (21^(3/2)+18^(3/2)-2*17^(3/2))=10.8053

    Which is not the same as what I got earlier: 18 Sqrt[2] + 7 Sqrt[21] = 57.5339

    What went wrong? Was it the discontinuity at x=0 that caused it?

     – Onizuka Oct 05 '14 at 00:07