Question

\[\int\limits_0^4 \left( x + e^{2x} \right) dx\]

Answer 1

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

\[\text{Here, }a = 0, b = 4, f\left( x \right) = x + e^{2x} , h = \frac{4 - 0}{n} = \frac{4}{n}\]
Therefore,
\[I = \int_0^4 \left( x + e^{2x} \right) d x\]
\[ = \lim_{h \to 0} h\left[ f\left( 0 \right) + f\left( 0 + h \right) + . . . . . . . . . . . . . . . . . . . . + f\left\{ 0 + \left( n - 1 \right)h \right\} \right]\]
\[ = \lim_{h \to 0} h\left[ \left( 0 + e^0 \right) + \left( h + e^{2h} \right) + . . . . . . . . . . . . . . . + \left\{ \left( n - 1 \right)h + e^{2\left( n - 1 \right)h} \right\} \right]\]
\[ = \lim_{h \to 0} h\left[ h\left\{ 1 + 2 + . . . . . + \left( n - 1 \right)h \right\} + e^0 + e^{2h} + e^{4h} + . . . . . . . . . + e^{2\left( n - 1 \right)h} \right]\]
\[ = \lim_{h \to 0} h\left[ h\frac{n\left( n - 1 \right)}{2} + \frac{\left( e^{2h} \right)^n - 1}{e^{2h} - 1} \right]\]
\[ = \lim_{n \to \infty} \frac{16}{n^2} \times \frac{n\left( n - 1 \right)}{2} + \lim_{h \to 0} \frac{e^8 - 1}{\frac{e^{2h} - 1}{h}}\]
\[ = \lim_{n \to \infty} 8\left( 1 - \frac{1}{n} \right) + \lim_{h \to 0} \frac{e^8 - 1}{\frac{2( e^{2h} - 1)}{2h}}\]
\[ = 8 + \frac{e^8 - 1}{2}\]
\[ = \frac{15 + e^8}{2}\]