Advertisements
Advertisements
Question
\[\int\limits_1^3 \left( x^2 + 3x \right) dx\]
Solution
\[\text{Here }a = 1, b = 3, f\left( x \right) = x^2 + 3x, h = \frac{3 - 1}{n} = \frac{2}{n}\]
Therefore,
\[ \int_1^3 \left( x^2 + 3x \right) d x = \lim_{h \to 0} h\left[ f\left( a \right) + f\left( a + h \right) + f\left( a + 2h \right) + . . . . . . . . . . . . + f\left( a + \left( n - 1 \right)h \right) \right]\]
\[ = \lim_{h \to 0} h\left[ f\left( 1 \right) + f\left( 1 + h \right) + . . . . . . . . . . + f\left( 1 + \left( n - 1 \right)h \right) \right]\]
\[ = \lim_{h \to 0} h\left[ 1 + 3 + \left( 1 + h \right)^2 + 3\left( 1 + h \right) + \left( 1 + 2h \right)^2 + 3\left( 1 + 2h \right) + . . . . . . . . . + \left( \left( n - 1 \right)h \right)^2 + 3\left( \left( n - 1 \right)h \right) \right]\]
\[ = \lim_{h \to 0} h\left[ n + h^2 \left( 1^2 + 2^2 + . . . . . . . . . . . . . . \left( n - 1 \right)^2 \right) + 2h\left( 1 + 2 + . . . . . . . . . . . . \left( n - 1 \right) \right) + 3n + 3h\left( 1 + 2 + . . . . . . . . . . . . \left( n - 1 \right) \right) \right]\]
\[ = \lim_{h \to 0} h\left[ 4n + h^2 \frac{n\left( n - 1 \right)\left( 2n - 1 \right)}{6} + 5h\frac{n\left( n - 1 \right)}{2} \right]\]
\[ = \lim_{n \to 0 } \left[ 8 + \frac{4}{3}\left( 1 - \frac{1}{n} \right)\left( 2 - \frac{1}{n} \right) + 10\left( 1 - \frac{1}{n} \right) \right]\]
\[ = 8 + \frac{8}{3} + 10\]
\[ = \frac{62}{3}\]
APPEARS IN
RELATED QUESTIONS
Evaluate the following definite integrals:
If \[\int\limits_0^a 3 x^2 dx = 8,\] write the value of a.
If \[\left[ \cdot \right] and \left\{ \cdot \right\}\] denote respectively the greatest integer and fractional part functions respectively, evaluate the following integrals:
\[\int\limits_0^\infty \frac{1}{1 + e^x} dx\] equals
The derivative of \[f\left( x \right) = \int\limits_{x^2}^{x^3} \frac{1}{\log_e t} dt, \left( x > 0 \right),\] is
\[\int\limits_0^{2a} f\left( x \right) dx\] is equal to
\[\int\limits_0^1 \tan^{- 1} \left( \frac{2x}{1 - x^2} \right) dx\]
\[\int\limits_0^{\pi/2} \frac{\sin x}{\sqrt{1 + \cos x}} dx\]
\[\int\limits_0^{\pi/4} \sin 2x \sin 3x dx\]
\[\int\limits_0^1 \left( \cos^{- 1} x \right)^2 dx\]
\[\int\limits_1^2 \frac{x + 3}{x\left( x + 2 \right)} dx\]
\[\int\limits_1^3 \left| x^2 - 4 \right| dx\]
\[\int\limits_{- \pi/4}^{\pi/4} \left| \tan x \right| dx\]
\[\int\limits_0^{15} \left[ x^2 \right] dx\]
\[\int\limits_0^\pi \cos 2x \log \sin x dx\]
\[\int\limits_0^\pi \frac{x \tan x}{\sec x + \tan x} dx\]
Using second fundamental theorem, evaluate the following:
`int_0^1 "e"^(2x) "d"x`
Evaluate the following using properties of definite integral:
`int_(- pi/2)^(pi/2) sin^2theta "d"theta`
Evaluate the following using properties of definite integral:
`int_0^(i/2) (sin^7x)/(sin^7x + cos^7x) "d"x`
Choose the correct alternative:
`Γ(3/2)`
Evaluate `int (x^2 + x)/(x^4 - 9) "d"x`
If `int (3"e"^x - 5"e"^-x)/(4"e"6x + 5"e"^-x)"d"x` = ax + b log |4ex + 5e –x| + C, then ______.
`int "e"^x ((1 - x)/(1 + x^2))^2 "d"x` is equal to ______.
