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:<math>\int^2_{-2}\sqrt{9t^4+16t^2} \, dt = \int^2_{-2} |t| \; \sqrt{9t^2+16} \, dt</math>
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dato che:
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:<math>d\left(9t^2+16\right) = 18t \, dt</math>
allora:
:<math>\int^2_{0}\frac{1}{18}\sqrt{9t^2+16} \ d\left(9t^2+16\right) \qquad \mathrm{per} \ t >0</math>
:<math>-\int^0_{-2}\frac{1}{18}\sqrt{9t^2+16} \ d\left(9t^2+16\right) \qquad \mathrm{per} \ t <0</math>
che è come se fosse l'integrale di:
:<math>\int \frac{1}{18}\sqrt{x} \, d\left(x\right)</math>
quindi:
:<math>\int^2_{-2}\frac{1}{18}\sqrt{9t^2+16} \ d\left(9t^2+16\right) = \frac{1}{18} \cdot \frac{2}{3} \left[ \left( 9t^2+16 \right)^{\frac{3}{2}} \right]^2_{0} - \frac{1}{18} \cdot \frac{2}{3} \left[ \left( 9t^2+16 \right)^{\frac{3}{2}} \right]^0_{-2} = \frac{1}{27} \left[ \left( 9t^2+16 \right)^{\frac{3}{2}} \right]^2_0 - \frac{1}{27} \left[ \left( 9t^2+16 \right)^{\frac{3}{2}} \right]^0_{-2} = ... </math>
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