Find the value of - n - 0 1 e t 2 -t 2t 2 -t-1 dt

∫ ^1_0e^t^2+t(2t^2+t+1)d t=∫ ^1_0 1.e^t^2+td t+∫ ^1_0 e^t^2+t (2 t^2+t) d t=[t.e^t^2+t]^1_0 ∫ ^1_0 e^t^2+t(2t+1)(t) dt +∫^1_0 e^t^2+t(2t^2+t)dt=e^2 ln ...

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Byju's Answer

Standard XII

Mathematics

Algebra of Limits

Find the valu...

Question

Find the value of $ℓ n (\int_{0}^{1} e^{t^{2} + t} (2 t^{2} + t + 1) d t)$

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Solution

$\int_{0}^{1} e^{t^{2} + t} (2 t^{2} + t + 1) d t = \int_{0}^{1} 1. e^{t^{2} + t} d t + \int_{0}^{1} e^{t^{2} + t} (2 t^{2} + t) d t = [t . e^{t^{2} + t}]_{0}^{1} - \int_{0}^{1} e^{t^{2} + t} (2 t + 1) (t) d t + \int_{0}^{1} e^{t^{2} + t} (2 t^{2} + t) d t = e^{2}$
$ln (\int_{0}^{1} e^{t^{2} + t} (2 t^{2} + t + 1) d t = ln (e^{2}) = 2$

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Find the value of ℓn(∫10et2+t(2t2+t+1)dt)

∫10et2+t(2t2+t+1)dt=∫101.et2+tdt+∫10et2+t(2t2+t)dt=[t.et2+t]10−∫10et2+t(2t+1)(t)dt+∫10et2+t(2t2+t)dt=e2ln(∫10et2+t(2t2+t+1)dt=ln(e2)=2

Find the value of $ℓ n (\int_{0}^{1} e^{t^{2} + t} (2 t^{2} + t + 1) d t)$