Question

Evaluate the following integrals as limit of sums:

\[\int_1^3 \left( 3 x^2 + 1 \right)dx\]

Answer 1

We have,

\[\int_a^b f\left( x \right)dx = \lim_{h \to 0} \left\{ f\left( a \right) + f\left( a + h \right) + f\left( a + 2h \right) + . . . + f\left[ \left( a + \left( n - 1 \right)h \right) \right] \right\}\]

Here, a = 1, b = 3,  f(x) = 3x² + 1 and

\[h = \frac{3 - 1}{n} = \frac{2}{n} \Rightarrow nh = 2\]

\[\therefore \int_1^3 \left( 3 x^2 + 1 \right)dx\]
\[ = \lim_{h \to 0} h \left\{ f\left( 1 \right) + f\left( 1 + h \right) + f\left( 1 + 2h \right) + . . . + f\left[ 1 + \left( n - 1 \right)h \right] \right\}\]
\[ = \lim_{h \to 0} h \left\{ \left[ 3 \times 1^2 + 1 \right] + \left[ 3 \times \left( 1 + h \right)^2 + 1 \right] + \left[ 3 \times \left( 1 + 2h \right)^2 + 1 \right] + . . . + \left[ 3 \times \left( 1 + \left( n - 1 \right)h \right)^2 + 1 \right] \right\}\]
\[ = \lim_{h \to 0} h\left\{ 3\left[ 1 + \left( 1 + 2h + h^2 \right) + \left( 1 + 4h + 2^2 h^2 \right) + . . . + \left( 1 + 2\left( n - 1 \right)h + \left( n - 1 \right)^2 h^2 \right) \right] + n \right\}\]
\[ = \lim_{h \to 0} h\left\{ 3\left[ n + 2\left( 1 + 2 + . . . + \left( n - 1 \right) \right)h + \left( 1^2 + 2^2 + . . . + \left( n - 1 \right)^2 \right) h^2 \right] + n \right\}\]
\[ = \lim_{h \to 0} h\left[ 4n + 6 \times \frac{n\left( n - 1 \right)}{2}h + 3 \times \frac{\left( n - 1 \right)n\left( 2n - 1 \right)}{6} h^2 \right]\]
\[= \lim_{h \to 0} \left[ 4nh + 6 \times \frac{nh\left( nh - h \right)}{2} + 3 \times \frac{\left( nh - h \right)nh\left( 2nh - h \right)}{6} \right]\]
\[ = \lim_{h \to 0} \left[ 4nh + 3 \times nh\left( nh - h \right) + 3 \times \frac{\left( nh - h \right)nh\left( 2nh - h \right)}{6} \right]\]
\[ = \lim_{h \to 0} \left[ 4 \times 2 + 3 \times 2 \times \left( 2 - h \right) + 3 \times \frac{\left( 2 - h \right) \times 2 \times \left( 2 \times 2 - h \right)}{6} \right]\]
\[ = 8 + 6 \times \left( 2 - 0 \right) + \frac{\left( 2 - 0 \right) \times 2 \times \left( 4 - 0 \right)}{2}\]
\[ = 8 + 12 + 8\]
\[ = 28\]