Loading Document√a2+u2= atanθFind :∫(x2+1)23dxRewrite :x=atanθ=tanθdx=sec2θdθSubstitute :∫(x2+1)23dx=∫(√tan2θ+1)3)sec2θdθ=∫(√secθ)3sec2θdθ=∫(secθ)3sec2θdθ=∫sec3θsec2θdθ=∫secθdθ=∫cosθ1dθ=∫cosθdθ= sinθ+c=√x2+1x+cEvaluate Definite Integral using Limit Definiton withReiimann sums..∫abf(x)dx=x→∞limi=1∑nf(Xi)ΔΧEvaluate:∫24(2x+5)dxStep 1.ΔΧ=n b−a=n 4−2=n2Step 2.Χi=a+iΔΧ=2+i(n2)Step 3.x→∞limi=1∑nf(2+n2i)⋅(n2i)x→∞limi=1∑n[2(2+n2i)+5] n2x→∞limi=1∑n(4+n4i+5) n2x→∞limi=1∑n(9+n4i) n2x→∞lim(n2i=1∑n9+n2i=1∑nn4i)x→∞lim(n18i=1∑n1+n28i=1∑ni)x→∞lim(n18(n)+n28(2n(n+1)))x→∞lim(18+n4(n+1))x→∞lim(18+n4n+4)x→∞lim(18+4+n4)Step 4.∫24(2x+5)dx=x→∞lim(22+n4)=22