Solved Problems Using Definition of Definite Integrals

Here we solve few problems of definite integrals.

Problem 1:  By the definition of definite integrals, find `int_a^b dx`.

Solution: From the definition of definite integrals, we have

`int_a^b dx=lim_{h to 0} h sum_{1}^{n} 1, quad` where ` nh=b-a`

            `=lim_{h to 0} h (1+1+1+ cdots text{till} n text{-terms})`

            `=lim_{h to 0} h cdot n`

            `=lim_{h to 0} b-a = b-a`

Problem 2:  By the definition of definite integrals, find `int_0^1 x dx`.

Solution: From the definition of definite integrals, we have

`int_0^1 x dx=lim_{h to 0} h sum_{r=1}^{n} rh, quad` where ` nh=1-0=1`

            `=lim_{h to 0} h (h+2h+3h+ cdots nh)`

            `=lim_{h to 0} h^2 (1+2+3+ cdots n)`

            `=lim_{h to 0} h^2 frac{n(n+1)}{2}`

[Formula used: `1+2+cdots+n=frac{n(n+1)}{2}`]

             `=lim_{h to 0}frac{nh(nh+h)}{2}`

            `=lim_{h to 0}frac{1(1+h)}{2}`

            `=frac{1(1+0)}{2}=frac{1}{2}`

Problem 3:  By the definition of definite integrals, find `int_1^2 x dx`.

Solution: From the definition of definite integrals, we have

`int_1^2 x dx=lim_{h to 0} h sum_{r=1}^{n} (1+rh), quad` where ` nh=2-1=1`

`=lim_{h to 0} h [(1+h)+(1+2h)+cdots+(1+nh)]`

`=lim_{h to 0} h [(1+1+ cdots text{till} n text{-terms})+` `h(1+2+cdots n)]`

`=lim_{h to 0}   h[n+h frac{n(n+1)}{2}]`

[Formula used: `1+2+cdots+n=frac{n(n+1)}{2}`]

`=lim_{h to 0}[nh+frac{nh(nh+h)}{2}]`

`=lim_{h to 0}[1+frac{1(1+h)}{2}] quad`as ` nh=1`

`=1+frac{1(1+0)}{2}=frac{3}{2}`

Problem 4:  By the definition of definite integrals, find `int_0^1 3x^2 dx`.

Solution: From the definition of definite integrals, we have

`int_0^1 3x^2 dx=lim_{h to 0} h sum_{r=1}^{n} 3(rh)^2, quad` where ` nh=1-0=1`

`=lim_{h to 0} h cdot 3h^2 sum_{r=1}^{n} r^2`

`=lim_{h to 0} 3h^3 (1^2+2^2+3^2+ cdots n^2)`

`=lim_{h to 0} 3h^3 cdot frac{n(n+1)(2n+1)}{6}`

[Formula used: `1^2+2^2+cdots+n^2=frac{n(n+1)(2n+1)}{6}`]

`=3lim_{h to 0}frac{nh(nh+h)(2nh+h)}{6}`

`=3lim_{h to 0}frac{1(1+h)(2+h)}{6} quad` as ` nh=1`

`=3frac{1(1+0)(2+0)}{6}=1`

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