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Displacement via u-substitution

A particle moves along a line with velocity v(t)=t(t2+1)4v(t) = t(t^2+1)^4 (in meters per second), for t≥0t \geq 0.

Find the displacement of the particle from t=0t=0 to t=2t=2, that is, evaluate ∫02t(t2+1)4 dt\int_0^2 t(t^2+1)^4 \, dt using the substitution u=t2+1u = t^2+1.

Hint

Let u=t2+1u = t^2+1, so du=2t dtdu = 2t\,dt, which means t dt=12dut\,dt = \frac{1}{2}du. Rewrite the bounds in terms of uu: when t=0t=0, u=1u=1; when t=2t=2, u=5u=5. Then integrate 12u4\frac{1}{2}u^4 with respect to uu over the new bounds.

Step-by-step solution

Substitute u=t2+1u = t^2+1, so du=2t dtdu = 2t\,dt, giving t dt=12dut\,dt = \frac{1}{2}du. Convert the bounds: at t=0t=0, u=02+1=1u = 0^2+1 = 1; at t=2t=2, u=22+1=5u = 2^2+1 = 5. The integral becomes ∫1512u4 du\int_1^5 \frac{1}{2}u^4\,du.

Evaluating, ∫1512u4 du=12⋅u55∣15=u510∣15\int_1^5 \frac{1}{2}u^4\,du = \frac{1}{2}\cdot\frac{u^5}{5}\Big|_1^5 = \frac{u^5}{10}\Big|_1^5. Plugging in gives 5510−1510=312510−110=312410=312.4\frac{5^5}{10} - \frac{1^5}{10} = \frac{3125}{10} - \frac{1}{10} = \frac{3124}{10} = 312.4. So the displacement is 312.4312.4 meters.

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