Mathematics

Evaluate the following integral as limit of sum:
$$\displaystyle \int_{1}^{4}(x^2-x)dx$$


SOLUTION
We know that
$$\displaystyle \int_a^b f(x) dx= \displaystyle \lim _{h\to 0} h[f(a)+f(a+h)+f(a+2h) +....+f(a+(n-1)h)]$$
Here $$a=1, b=4, h=\dfrac {3}{n} $$ and $$f(x) =x^2 -x$$
$$\therefore \ I=\displaystyle \int_{1}^4 (x^2-x) dx \displaystyle \lim_{h\to 0} h[ f(1) + f(1+h)+f(1-2h) +...+f(1+(n-1) h)]$$
$$\Rightarrow =I\displaystyle \lim _{ h\rightarrow 0 }{ h\left[ \left\{ \left( { 1 }^{ 2 }-1 \right)  \right\} +\left\{ { \left( 1+h \right)  }^{ 2 }-\left( 1+h \right)  \right\} +\left\{ { \left( 1+2h \right)  }^{ 2 }-\left( 1+2h \right)  \right\} +...+\left\{ { \left( 1+\left( n-1 \right) h \right)  }^{ 2 }-\left( 1+\left( n-1 \right) h \right)  \right\}  \right]  } $$
$$\Rightarrow =I\displaystyle \lim _{ h\rightarrow 0 }{ h\left[ \left\{ { 1 }^{ 2 }+{ \left( 1+h \right)  }^{ 2 }+...+{ \left( 1+\left( n-1 \right) h \right)  }^{ 2 } \right\} -\left\{ 1+\left( 1+h \right) +\left( 1+2h \right) +....+\left( 1+\left( n-1 \right) h \right)  \right\}  \right]  } $$
$$\Rightarrow =I\displaystyle \lim _{ h\rightarrow 0 }{ h\left[ \left\{ 1+2h\left( 1+2+3+...+\left( n-1 \right) h \right) +{ h }^{ 2 }\left( { 1 }^{ 2 }+{ 2 }^{ 2 }+....+{ \left( n-1 \right)  }^{ 2 } \right)  \right\} \left\{ n+h\left( 1+2+3+...+\left( n-1 \right)  \right)  \right\}  \right]  } $$
$$\Rightarrow =I\displaystyle \lim _{ h\rightarrow 0 }{ h\left[ \left\{ n+hn\left( n-1 \right) +{ h }^{ 2 }\frac { n\left( n-1 \right) \left( 2n-1 \right)  }{ 6 }  \right\} -\left\{ n+h\frac { n\left( n-1 \right)  }{ 2 }  \right\}  \right]  } $$
$$\Rightarrow \ I=\displaystyle \lim_{h\to 0}h \left [h\dfrac {(n-1)}{2}+h^2 \dfrac {n(n-1) (2n-1)}{n^2}\right]$$
$$\Rightarrow \ I=\displaystyle \lim_{h\to \infty} \left\{\dfrac {9}{2} \left (\dfrac {n-1}{n}\right) +\dfrac {27}{6} \dfrac {(n-1)(2n-1)}{n^2}\right\}$$
$$\Rightarrow \ I=\displaystyle \lim_{h\to \infty} \left\{\dfrac {9}{2} \left(1-\dfrac {1}{n}\right) +\dfrac {9}{2} \left (2-\dfrac {1}{n}\right) \right\} =\dfrac {9}{2} +9=\dfrac {27}{2}$$

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