Question

Calculus (2) Master Needed to show steps of evaluating the integral b=0, a= pi/2 :$\int\limits^a_b {(t^2-3t)sint} \, dt$ I get stuck after applying integration by parts and simplified the negative, I get $(t^2-3t)*-cost |\frac{\pi /2}{0} + \int\limits^a_b {cost * (t-3)} \, dt$ what am I supposed to do at this stage? Or did I do it wrong? Step by step is needed!

Answer 1

Answer:

$\displaystyle \int\limits^{\pi/2}_ {0} \,(t^2-3t)\sin t\, dt = \pi -5$

Step-by-step explanation:

We want to evaluate the integral:

$\displaystyle \int\limits^{\pi/2}_ {0} \,(t^2-3t)\sin(t)\, dt$

Since this is a product of two functions, we can use Integration by Parts. Recall that:

$\displaystyle \int u\, dv = uv - \int v\, du$

Let u be t² - 3t and let dv be sin(t).

Thus:

$du = 2t - 3\, dt$

And:

$\displaystyle v = \int \sin t\, dt = - \cos t$

So:

$\displaystyle =\left(-\cos t\right)\left(t^2-3t\right)\Bigg|_0^{\pi/2}-\int_{0}^{\pi/2} \left(-\cos t\right)\left(2t-3\right)\, dt$

Simplify:

$\displaystyle = - \left(\cos t\right)(t^2-3t)\Bigg|_{0}^{\pi /2}+\int\limits^{\pi/2}_ {0} \,\cos(t)(2t-3)\, dt$

We can use IBP once more.

Let u be 2t - 3 and let dv be cos(t). Therefore:

$du = 2\, dt$

And:

$\displaystyle v = \int \cos t\, dt = \sin t$

So:

$\displaystyle = - \left(\cos t\right)(t^2-3t)\Bigg|_0^{\pi/2}+\left(\sin(t)(2t-3)\Bigg|_0^{\pi /2}-\int\limits^{\pi/2}_0 \, 2\sin(t)dt \right)$

Evaluate:

$\displaystyle =-\left(\cos t\right)(t^2-3t)+\sin(t)(2t-3)+2\cos(t)\Bigg|_0^{\pi /2}$

Evaluate the first section:

$\displaystyle \begin{aligned} &= -\cos\left(\frac{\pi}{2}\right)\left(\left(\frac{\pi}{2}\right)^2-3\left(\frac{\pi}{2}\right)\right)+\sin\left(\frac{\pi}{2}\right)\left(2\left(\frac{\pi}{2}\right)-3\right)+2\cos \left(\frac{\pi}{2}\right)\right) \\ \\ &= (0) + (1)(\pi -3) + (0) \\ \\ &= \pi -3 \end{aligned}$

And the second:

$\displaystyle \begin{aligned} &= -\cos(0)((0)^2-3(0))+(\sin(0))(2(0)-3)+2\cos(0 ) \\ \\&=(0)+(0)+(2) \\ \\ &= 2\end{aligned}$

Therefore:

$\displaystyle \int\limits^{\pi/2}_ {0} \,(t^2-3t)\sin(t)\, dt = (\pi -3) -(2) = \pi -5$