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Exercise 4.1 · Q2

Q.Evaluate the following integral as limit of sum: ∫04x2 dx\int_0^4 x^2\,dx

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With f(x)=x2f(x)=x^2, h=4/nh=4/n, f(rh)=r2h2f(rh)=r^2h^2.

∑r=1nh⋅r2h2=h3∑r2=h3⋅n(n+1)(2n+1)6=(nh)36(1+1n)(2+1n).\sum_{r=1}^n h\cdot r^2h^2 = h^3\sum r^2 = h^3\cdot\frac{n(n+1)(2n+1)}6=\frac{(nh)^3}{6}\Big(1+\frac1n\Big)\Big(2+\frac1n\Big).

As n→∞n\to\infty, nh=4nh=4 fixed, so the limit is 646(1)(2)=1286=643\dfrac{64}{6}(1)(2)=\dfrac{128}{6}=\dfrac{64}{3}.

✓Final answer

∫04x2 dx=643\displaystyle\int_0^4 x^2\,dx=\dfrac{64}{3}

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