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Exercise 4.1 · Q1

Q.Evaluate the following integral as limit of sum: ∫13(3x−4) dx\int_1^3 (3x-4)\,dx

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✓ Free question

With f(x)=3x−4f(x)=3x-4, a=1a=1, b=3b=3, h=2nh=\dfrac{2}{n}, so nh=2nh=2 and f(a+rh)=3(1+rh)−4=−1+3rhf(a+rh)=3(1+rh)-4=-1+3rh.

∑r=1nh f(a+rh)=h∑r=1n(−1+3rh)=−hn+3h2⋅n(n+1)2=−(hn)+32(hn)2(1+1n).\sum_{r=1}^n h\,f(a+rh)=h\sum_{r=1}^n(-1+3rh)=-hn+3h^2\cdot\frac{n(n+1)}2=-(hn)+\frac32(hn)^2\Big(1+\frac1n\Big).

As n→∞n\to\infty: hn→2hn\to2, 1n→0\frac1n\to0, so the limit is −2+32(4)(1)=−2+6=4-2+\frac32(4)(1)=-2+6=4.

✓Final answer

∫13(3x−4) dx=4\displaystyle\int_1^3(3x-4)\,dx = 4

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