Question

Evaluate the integral as limit of a sum: $\int\limits^3_1 {x^{3}} \, dx$

Answer 1

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Answer 2

Answer:

$\int\limits^3_1 {x^{3}} \, dx=20$

Step-by-step explanation:

To evaluate the integral as a limit of sum

here a=1, b = 3 , nh=b-a =2

$\int\limits^a_b {f(x)} \, dx = \lim_{h \to 0} h[f(a)+f(a+h)+f(a+2h)+...+f(a+(n-1)h)] \\\\\\\int\limits^1_3 {x^{3}} \, dx = \lim_{n \to \infty} h[1^{3}+(1+h)^{3}+(1+2h)^{3}+...+(1+(n-1)h)^{3}] \\\\\\=\lim_{h \to 0} h[1+1+h^{3}+3h^{2}+3h+1+8h^{3}+12h^{2}+6h+...+1+(n-1)^{3}h^{3}]$

$\lim_{h \to 0} h[1+1+1+...+(n\:times)+h(3+6+9+...+(n-1)+h^{2}(3+6+12+...+(n-1)+h^{3}(1+8+27+...+(n-1)]\\\\=\lim_{h \to 0}h[n+h\frac{n(n-1)}{2} +h^{2}\frac{n(n-1)(2n-1)}{6}+h^{3}(\frac{n(n+1)}{2} )^{2}]$

now put limits and value of nh

$\int\limits^3_1 {x^{3}} \, dx=20$