Solution :
The velocity distribution in boundary layer given by
Uu=2(δy)−(δy)2
1. Displacement thickness ( δ∗)
δ∗=∫0δ(1−Uu)dy
Substituting Uu=2(δy)−(δy)2 in above equation, we get
δ∗=∫0δ{1−[2(δy)−(δy)2]}dy=∫0δ{1−2(δy)+(δy)2}dy=[y−2δ2y2+3δ2y3]0δ=δ−δδ2+3δ2δ3=δ−δ+3δ=3δ
2. Momentum thickness (θ)
θ=∫0δUu{1−Uu}dy=∫0δ(δ2y−δ2y2)[1−(δ2y−δ2y2)]dy=∫0δ[δ2y−δ2y2][1−δ2y+δ2y2]dy=∫0δ[δ2y−δ24y2+δ32y3−δ2y2+δ32y3−δ4y4]dy=∫0δ[δ2y−δ25y2+δ34y3−δ4y4]dy=[2δ2y2−3δ25y3+4δ34y4−5δ4y5]0δ=[δδ2−3δ25δ3+δ3δ4−5δ4δ5]=δ−35δ+δ−5δ=1515δ−25δ+15δ−3δ=1530δ−28δ=152δ
3. Energy thickness θ∗∗
δ∗∗=∫0δUu[1−U2u2]dy=∫0δ(δ2y−δ2y2)(1−[δ2y−δ2y2]2)dy=∫0δ(δ2y−δ2y2)(1−[δ24y2+δ4y4−δ34y3])dy=∫0δ(δ2y−δ2y2)(1−δ24y2−δ4y4+δ34y3)dy=∫0δ(δ2y−δ38y3−δ52y5+δ48y4−δ2y2+δ44y4+δ6y6−δ54y5)dy=∫00[δ2y−δ2y2−δ38y3+δ412y4−δ36y5+δ6y6]dy=[2δ2y2−3δ2y3−4δ38y4+5δ412y5−6δ56y6+7δ6y7]0δ=δδ2−3δ2δ3−δ32δ4+5δ412δ5−δ5δ6+7δ6δ7=δ−3δ−2δ+512δ−δ+7δ=−2δ−3δ+512δ+7δ=105−210δ−35δ+252δ+15δ=105−245δ+267δ=10522δ
we calculated Displacement thickness, momentum thickness and energy thickness.