Question Number 70074 by cat2315 last updated on 30/Sep/19
![∫_1 ^2 [3+(1/t^2 )]dt=](Q70074.png)
$$\int_{\mathrm{1}} ^{\mathrm{2}} \left[\mathrm{3}+\frac{\mathrm{1}}{{t}^{\mathrm{2}} }\right]{dt}= \\ $$
Commented by mathmax by abdo last updated on 30/Sep/19
![if [..]means the integr part we get ∫_1 ^2 [3+(1/t^2 )]dt =∫_1 ^2 (3+[(1/t^2 )])dt =3 +∫_1 ^2 [(1/t^2 )]dt for 1≤t≤2 ⇒(1/2)≤(1/t)≤1 ⇒(1/4) ≤(1/t^2 ) ≤1 ⇒[(1/t^2 )]=0 ⇒ ∫_1 ^2 [3+(1/t^2 )]dt =3](Q70101.png)
$${if}\:\left[..\right]{means}\:{the}\:{integr}\:{part}\:\:{we}\:{get} \\ $$$$\int_{\mathrm{1}} ^{\mathrm{2}} \left[\mathrm{3}+\frac{\mathrm{1}}{{t}^{\mathrm{2}} }\right]{dt}\:=\int_{\mathrm{1}} ^{\mathrm{2}} \left(\mathrm{3}+\left[\frac{\mathrm{1}}{{t}^{\mathrm{2}} }\right]\right){dt}\:=\mathrm{3}\:+\int_{\mathrm{1}} ^{\mathrm{2}} \left[\frac{\mathrm{1}}{{t}^{\mathrm{2}} }\right]{dt} \\ $$$${for}\:\mathrm{1}\leqslant{t}\leqslant\mathrm{2}\:\Rightarrow\frac{\mathrm{1}}{\mathrm{2}}\leqslant\frac{\mathrm{1}}{{t}}\leqslant\mathrm{1}\:\Rightarrow\frac{\mathrm{1}}{\mathrm{4}}\:\leqslant\frac{\mathrm{1}}{{t}^{\mathrm{2}} }\:\leqslant\mathrm{1}\:\Rightarrow\left[\frac{\mathrm{1}}{{t}^{\mathrm{2}} }\right]=\mathrm{0}\:\Rightarrow \\ $$$$\int_{\mathrm{1}} ^{\mathrm{2}} \left[\mathrm{3}+\frac{\mathrm{1}}{{t}^{\mathrm{2}} }\right]{dt}\:=\mathrm{3} \\ $$
Answered by Rio Michael last updated on 30/Sep/19
![∫_1 ^2 [3+ t^(−2) ]dt = [3t − t^(−1) ]_1 ^2 = [3t −(1/t)]_1 ^2 = [(6 −(1/2)) − 2] = (7/2)](Q70086.png)
$$\int_{\mathrm{1}} ^{\mathrm{2}} \left[\mathrm{3}+\:{t}^{−\mathrm{2}} \right]{dt}\:=\:\left[\mathrm{3}{t}\:\:−\:{t}^{−\mathrm{1}} \right]_{\mathrm{1}} ^{\mathrm{2}} \\ $$$$\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:=\:\:\left[\mathrm{3}{t}\:−\frac{\mathrm{1}}{{t}}\right]_{\mathrm{1}} ^{\mathrm{2}} \\ $$$$\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:=\:\left[\left(\mathrm{6}\:−\frac{\mathrm{1}}{\mathrm{2}}\right)\:−\:\mathrm{2}\right]\:=\:\frac{\mathrm{7}}{\mathrm{2}} \\ $$$$\:\:\:\: \\ $$