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I decided to use U-substitution to show that $s=61$ but have failed. Here is my working, where have I gone wrong? Regards Tom $$s=\int_0^33t\sqrt{t^2+16}\,dt$$ $$u=t^2+16$$ $$du=2tdt\Rightarrow3(\frac{du}{2})=3tdt$$ $$t=3\Rightarrow(3)^2+16=25$$ $$t=0\Rightarrow(0)^2+16=16$$ $$s=\int_{16}^{25}(\sqrt{u})(3(\frac{du}{2}))$$ $$s=\frac32\int_{16}^{25}(\sqrt{u})du$$ $$s=\frac32\int_{16}^{25}u^{1/2}du$$ $$s=\frac32\left[\frac{u^{3/2}}{\frac32}\right]_{16}^{25}$$ $$s=\frac32\left[\frac23u^{3/2}\right]_{16}^{25}$$ $$s=\frac32\left[\frac23(t^2+16)^{3/2}\right]_{16}^{25}$$ $$s=\frac32\left[\frac23((25)^2+16)^{3/2})\right]-\frac32\left[\frac23((16)^2+16)^{3/2})\right]$$

$$(=11742.884....)$$

Thomas Winkworth
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2 Answers2

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OK up to

$$ s=\frac32\left[\frac23u^{3/2}\right]_{16}^{25} $$

but then you have kept the revised limits of integration (for $u$) when you revert back to $t$ as the variable:

$$ s=\frac32\left[\frac23(t^2+16)^{3/2}\right]_{16}^{25} $$

David Holden
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Your flaw is in substituting $t$ back in. For one thing, you don't need to do this, and for another, you didn't change the limits back. If you go $$ s= \frac{3}{2} \left[ \frac{2}{3}u^{3/2} \right]_{16}^{25} = [u^{3/2}]_{16}^{25}, $$ then you get $$ 25^{3/2} - 16^{3/2} = 5^3-4^3 = 125-64=61. $$

Chappers
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