Advertisements
Advertisements
प्रश्न
\[\int\limits_0^{15} \left[ x^2 \right] dx\]
Advertisements
उत्तर
We have,
\[I = \int\limits_0^{1 . 5} \left[ x^2 \right] dx\]
\[ = \int\limits_0^1 \left[ x^2 \right] dx + \int\limits_1^\sqrt{2} \left[ x^2 \right] dx + \int\limits_\sqrt{2}^{1 . 5} \left[ x^2 \right] dx\]
\[ = \int\limits_0^1 \left( 0 \right) dx + \int\limits_1^\sqrt{2} \left( 1 \right) dx + \int\limits_\sqrt{2}^{1 . 5} \left( 2 \right) dx ..............\left(\because \left[ x^2 \right] = \begin{cases}0 &where,& 0 < x < 1 \\ 1 &where,& 1 < x < \sqrt{2}\\2 &where,& \sqrt{2} < x < 1.5 \end{cases}\right)\]
\[ = 0 + \left[ x \right]_1^\sqrt{2} + \left[ 2x \right]_\sqrt{2}^{1 . 5} \]
\[ = \left[ x \right]_1^\sqrt{2} + 2 \left[ x \right]_\sqrt{2}^{1 . 5} \]
\[ = \left( \sqrt{2} - 1 \right) + 2\left( 1 . 5 - \sqrt{2} \right)\]
\[ = \sqrt{2} - 1 + 3 - 2\sqrt{2}\]
\[ = 2 - \sqrt{2}\]
APPEARS IN
संबंधित प्रश्न
\[\int\limits_{\pi/4}^{\pi/2} \cot x\ dx\]
\[\int\limits_0^{( \pi )^{2/3}} \sqrt{x} \cos^2 x^{3/2} dx\]
Evaluate the following integral:
Given that \[\int\limits_0^\infty \frac{x^2}{\left( x^2 + a^2 \right)\left( x^2 + b^2 \right)\left( x^2 + c^2 \right)} dx = \frac{\pi}{2\left( a + b \right)\left( b + c \right)\left( c + a \right)},\] the value of \[\int\limits_0^\infty \frac{dx}{\left( x^2 + 4 \right)\left( x^2 + 9 \right)},\]
\[\int\limits_0^{2a} f\left( x \right) dx\] is equal to
\[\int\limits_0^1 \cos^{- 1} x dx\]
\[\int\limits_0^1 \frac{1 - x}{1 + x} dx\]
\[\int\limits_0^{\pi/2} \frac{1}{1 + \tan^3 x} dx\]
\[\int\limits_0^{\pi/2} \frac{x}{\sin^2 x + \cos^2 x} dx\]
\[\int\limits_{- \pi}^\pi x^{10} \sin^7 x dx\]
\[\int\limits_{- 1}^1 e^{2x} dx\]
\[\int\limits_0^3 \left( x^2 + 1 \right) dx\]
Using second fundamental theorem, evaluate the following:
`int_0^1 "e"^(2x) "d"x`
Using second fundamental theorem, evaluate the following:
`int_1^"e" ("d"x)/(x(1 + logx)^3`
If x = `int_0^y "dt"/sqrt(1 + 9"t"^2)` and `("d"^2y)/("d"x^2)` = ay, then a equal to ______.
