Advertisements
Advertisements
प्रश्न
\[\int\limits_{- 1}^1 e^{2x} dx\]
Advertisements
उत्तर
\[\text{Here }a = - 1, b = 1, f\left( x \right) = e^{2x} , h = \frac{1 + 1}{n} = \frac{2}{n}\]
Therefore,
\[ \int_{- 1}^1 e^{2x} d x = \lim_{h \to 0} h\left[ f\left( a \right) + f\left( a + h \right) + f\left( a + 2h \right) + . . . . . . . . . . . . + f\left( a + \left( n - 1 \right)h \right) \right]\]
\[ = \lim_{h \to 0} h\left[ f\left( - 1 \right) + f\left( - 1 + h \right) + . . . . . . . . . . + f\left( - 1 + \left( n - 1 \right)h \right) \right]\]
\[ = \lim_{h \to 0} h\left[ e^{- 2} + e^{2\left( - 1 + h \right)} + e^{2\left( - 1 + 2h \right)} + . . . . . . . + e^{2\left( - 1 + \left( n - 1 \right)h \right)} \right]\]
\[ = \lim_{h \to 0} h e^{- 2} \left[ \frac{\left( e^{2h} \right)^n - 1}{e^{2h} - 1} \right]\]
\[ = \lim_{h \to 0} e^{- 2} \left[ \frac{e^4 - 1}{\frac{e^{2h} - 1}{2h}} \right] \times \frac{1}{2} .......................\left(\text{Since, nh = 2 }\right)\]
\[ = \frac{1}{2}\left( e^2 - e^{- 2} \right)\]
APPEARS IN
संबंधित प्रश्न
If \[\int_0^a \frac{1}{4 + x^2}dx = \frac{\pi}{8}\] , find the value of a.
\[\int\limits_0^{\pi/2} \frac{1}{2 + \cos x} dx\] equals
\[\int\limits_0^{2a} f\left( x \right) dx\] is equal to
The value of \[\int\limits_0^1 \tan^{- 1} \left( \frac{2x - 1}{1 + x - x^2} \right) dx,\] is
The value of \[\int\limits_{- \pi/2}^{\pi/2} \left( x^3 + x \cos x + \tan^5 x + 1 \right) dx, \] is
\[\int\limits_1^5 \frac{x}{\sqrt{2x - 1}} dx\]
\[\int\limits_0^1 \cos^{- 1} \left( \frac{1 - x^2}{1 + x^2} \right) dx\]
\[\int\limits_0^{\pi/2} \frac{\sin^2 x}{\left( 1 + \cos x \right)^2} dx\]
\[\int\limits_1^3 \left| x^2 - 4 \right| dx\]
\[\int\limits_0^{\pi/2} \frac{1}{1 + \cot^7 x} dx\]
\[\int\limits_0^a \frac{\sqrt{x}}{\sqrt{x} + \sqrt{a - x}} dx\]
\[\int\limits_0^{\pi/2} \frac{1}{1 + \tan^3 x} dx\]
\[\int\limits_0^4 x dx\]
\[\int\limits_0^2 \left( x^2 + 2 \right) dx\]
Evaluate the following using properties of definite integral:
`int_0^(i/2) (sin^7x)/(sin^7x + cos^7x) "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 ______.
Evaluate the following:
`int ((x^2 + 2))/(x + 1) "d"x`
`int x^3/(x + 1)` is equal to ______.
