Advertisements
Advertisements
प्रश्न
\[\int\limits_1^3 \left( 2 x^2 + 5x \right) dx\]
उत्तर
\[\text{Here, }a = 1, b = 3, f\left( x \right) = 2 x^2 + 5x, h = \frac{3 - 1}{n} = \frac{2}{n}\]
Therefore,
\[ \int_1^3 \left( 2 x^2 + 5x \right) d x = \lim_{h \to 0} h\left[ f\left( a \right) + f\left( a + h \right) + f\left( a + 2h \right) + . . . . . . . . . . . . + f\left( a + \left( n - 1 \right)h \right) \right]\]
\[ = \lim_{h \to 0} h\left[ f\left( 1 \right) + f\left( 1 + h \right) + . . . . . . . . . . + f\left( 1 + \left( n - 1 \right)h \right) \right]\]
\[ = \lim_{h \to 0} h\left[ 2 + 5 + 2 \left( 1 + h \right)^2 + 5\left( 1 + h \right) + 2 \left( 1 + 2h \right)^2 + 5\left( 1 + 2h \right) + . . . . . . . . . + 2 \left( \left( n - 1 \right)h \right)^2 + 5\left( \left( n - 1 \right)h \right) \right]\]
\[ = \lim_{h \to 0} h\left[ 2n + 2 h^2 \left( 1^2 + 2^2 + . . . . . . . . . . . . . . \left( n - 1 \right)^2 \right) + 4h\left( 1 + 2 + . . . . . . . . . . . . \left( n - 1 \right) \right) + 5n + 5h\left( 1 + 2 + . . . . . . . . . . . . \left( n - 1 \right) \right) \right]\]
\[ = \lim_{h \to 0} h\left[ 7n + 2 h^2 \frac{n\left( n - 1 \right)\left( 2n - 1 \right)}{6} + 9h\frac{n\left( n - 1 \right)}{2} \right]\]
\[ = \lim_{n \to 0 } \left[ 14 + \frac{8}{3}\left( 1 - \frac{1}{n} \right)\left( 2 - \frac{1}{n} \right) + 18\left( 1 - \frac{1}{n} \right) \right]\]
\[ = 14 + \frac{16}{3} + 18\]
\[ = \frac{112}{3}\]
APPEARS IN
संबंधित प्रश्न
Evaluate each of the following integral:
The value of the integral \[\int\limits_0^{\pi/2} \frac{\sqrt{\cos x}}{\sqrt{\cos x} + \sqrt{\sin x}} dx\] is
\[\int_0^\frac{\pi^2}{4} \frac{\sin\sqrt{x}}{\sqrt{x}} dx\] equals
Evaluate : \[\int\frac{dx}{\sin^2 x \cos^2 x}\] .
\[\int\limits_1^2 x\sqrt{3x - 2} dx\]
\[\int\limits_0^1 \cos^{- 1} \left( \frac{1 - x^2}{1 + x^2} \right) dx\]
Evaluate the following integrals :-
\[\int_2^4 \frac{x^2 + x}{\sqrt{2x + 1}}dx\]
\[\int\limits_0^{\pi/2} \frac{1}{1 + \cot^7 x} dx\]
\[\int\limits_0^{2\pi} \cos^7 x dx\]
\[\int\limits_0^\pi \frac{x \sin x}{1 + \cos^2 x} dx\]
\[\int\limits_0^\pi x \sin x \cos^4 x dx\]
\[\int\limits_0^{\pi/2} \frac{\sin^2 x}{\sin x + \cos x} dx\]
\[\int\limits_0^1 \cot^{- 1} \left( 1 - x + x^2 \right) dx\]
Evaluate the following integrals as the limit of the sum:
`int_0^1 (x + 4) "d"x`
Choose the correct alternative:
`int_(-1)^1 x^3 "e"^(x^4) "d"x` is
`int (cos2x - cos 2theta)/(cosx - costheta) "d"x` is equal to ______.
`int x^3/(x + 1)` is equal to ______.
`int (x + 3)/(x + 4)^2 "e"^x "d"x` = ______.
