Advertisements
Advertisements
प्रश्न
Advertisements
उत्तर
\[\int_a^b f\left( x \right) d x = \lim_{h \to 0} h\left[ f\left( a \right) + f\left( a + h \right) + f\left( a + 2h \right) . . . . . . . . . . . . . . . + f\left\{ a + \left( n - 1 \right)h \right\} \right]\]
\[\text{where }h = \frac{b - a}{n}\]
\[\text{Here }a = a, b = b, f\left( x \right) = x, h = \frac{b - a}{n}\]
Therefore,
\[I = \int_a^b x d x\]
\[ = \lim_{h \to 0} h\left[ f\left( a \right) + f\left( a + h \right) + . . . . . . . . . . . . . . . . . . . . + f\left\{ a + \left( n - 1 \right)h \right\} \right]\]
\[ = \lim_{h \to 0} h\left[ a + \left( a + h \right) + \left( a + 2h \right) + . . . . . . . . . . + \left\{ a + \left( n - 1 \right)h \right\} \right]\]
\[ = \lim_{h \to 0} h\left[ na + h\left\{ 1 + 2 + 3 + . . . . . . . . + \left( n - 1 \right) \right\} \right]\]
\[ = \lim_{h \to 0} h\left[ na + h\frac{n\left( n - 1 \right)}{2} \right]\]
\[ = \lim_{h \to 0} \frac{b - a}{n}\left[ na + \frac{\left[ b - a \right]\left( n - 1 \right)}{2} \right]\]
\[ = \lim_{h \to 0} \left[ \left( b - a \right)a + \frac{\left( b - a \right)\left( b - a - h \right)}{2} \right]\]
\[ = \left( b - a \right)a + \frac{\left( b - a \right)^2}{2}\]
\[ = \frac{2ab - 2 a^2 + b^2 + a^2 - 2ab}{2}\]
\[ = \frac{b^2 - a^2}{2}\]
APPEARS IN
संबंधित प्रश्न
\[\int\limits_1^4 f\left( x \right) dx, where f\left( x \right) = \begin{cases}7x + 3 & , & \text{if }1 \leq x \leq 3 \\ 8x & , & \text{if }3 \leq x \leq 4\end{cases}\]
Evaluate the following integral:
If f is an integrable function, show that
\[\int\limits_{- a}^a f\left( x^2 \right) dx = 2 \int\limits_0^a f\left( x^2 \right) dx\]
The value of \[\int\limits_0^\pi \frac{x \tan x}{\sec x + \cos x} dx\] is __________ .
If f (a + b − x) = f (x), then \[\int\limits_a^b\] x f (x) dx is equal to
The value of \[\int\limits_0^{\pi/2} \log\left( \frac{4 + 3 \sin x}{4 + 3 \cos x} \right) dx\] is
\[\int\limits_0^{\pi/2} \frac{\sin^2 x}{\left( 1 + \cos x \right)^2} dx\]
\[\int\limits_0^1 \left( \cos^{- 1} x \right)^2 dx\]
\[\int\limits_0^{15} \left[ x^2 \right] dx\]
\[\int\limits_0^{\pi/2} \frac{\cos^2 x}{\sin x + \cos x} dx\]
\[\int\limits_0^2 \left( x^2 + 2 \right) dx\]
Using second fundamental theorem, evaluate the following:
`int_1^2 (x "d"x)/(x^2 + 1)`
Using second fundamental theorem, evaluate the following:
`int_1^"e" ("d"x)/(x(1 + logx)^3`
Using second fundamental theorem, evaluate the following:
`int_1^2 (x - 1)/x^2 "d"x`
Evaluate the following:
f(x) = `{{:("c"x",", 0 < x < 1),(0",", "otherwise"):}` Find 'c" if `int_0^1 "f"(x) "d"x` = 2
Choose the correct alternative:
Γ(1) is
Verify the following:
`int (x - 1)/(2x + 3) "d"x = x - log |(2x + 3)^2| + "C"`
