Evaluate ∫31(2x²+5x)dx as a limit of a sum.

Letf(x)=2x2+5xHerea=1,b=3Therefore,h=b−an=3𕒵n=2n⇒nh=2Alson→∞⇔h→0We know that,∫abf(x)dx=limh→0h[f(a)+f(a+h)+...+f(a+(n𕒵)h)]∫31(2x2+5x)dx=limh→0h[f(1)+f(1+h)+...+f(1+(n𕒵)h)]=limh→0h[(2×12+5×1)+2(1+h)2+5(1+h)++2(1−(n𕒵)h2+5(1+(n𕒵)d))]=limh→0h[(2+5)+(2+4h+2h2+5+5h)++(2+4(n𕒵)h+2(n𕒵)2h2+5+5(n𕒵)h)]=limh→0h[(7)+(7+9h+2h2)++(7+9(n𕒵)h+2(n𕒵)2h2]=limh→0h[(7n+9h(1+2++n𕒵)+2h2(12+22+(n𕒵)2))]=limh→0[7nh+9h2n(n𕒵)2+2h3(n𕒵).n(2n𕒵)6]=limh→0⎡⎢⎢⎢⎢⎣7nh+9(nh)2 (1𕒵n)2+2(nh)3 (1𕒵n) (2𕒵n)6⎤⎥⎥⎥⎥⎦=limn→∞⎡⎢⎢⎢⎢⎣14+36(1𕒵n)2+16(1𕒵n) (2𕒵n)6⎤⎥⎥⎥⎥⎦[∴nh=2]=limn→∞[14+18(1𕒵n)+83(1𕒵n) (2𕒵n)]14+18+83×1×2=32+163=96+163=1123.