Advertisements
Advertisements
प्रश्न
If \[\left[ \cdot \right] and \left\{ \cdot \right\}\] denote respectively the greatest integer and fractional part functions respectively, evaluate the following integrals:
Advertisements
उत्तर
\[\text{We have}, \]
\[I = \int\limits_0^{\pi/4} \sin \left\{ x \right\} dx\]
\[\text{We know that}, \]
\[\left\{ x \right\} = x\text{, when }0 < x < \frac{\pi}{4} ..............\left(\text{As }\pi = 3 . 14 \Rightarrow \frac{\pi}{4} = 0 . 785 < 1 \right)\]
\[ \therefore I = \int\limits_0^{\pi/4} \sin x\ dx\]
\[ = \left[ - \cos x \right]_0^\frac{\pi}{4} \]
\[ = - \left( \cos \frac{\pi}{4} - \cos 0 \right)\]
\[ = \cos 0 - \cos \frac{\pi}{4}\]
\[ = 1 - \frac{1}{\sqrt{2}}\]
\[ = \frac{\sqrt{2} - 1}{\sqrt{2}}\]
APPEARS IN
संबंधित प्रश्न
`int_0^1 sqrt((1 - "x")/(1 + "x")) "dx"`
The derivative of \[f\left( x \right) = \int\limits_{x^2}^{x^3} \frac{1}{\log_e t} dt, \left( x > 0 \right),\] is
Evaluate : \[\int\limits_0^\pi/4 \frac{\sin x + \cos x}{16 + 9 \sin 2x}dx\] .
\[\int\limits_0^{\pi/4} \cos^4 x \sin^3 x dx\]
\[\int\limits_{\pi/3}^{\pi/2} \frac{\sqrt{1 + \cos x}}{\left( 1 - \cos x \right)^{5/2}} dx\]
\[\int\limits_{- 1/2}^{1/2} \cos x \log\left( \frac{1 + x}{1 - x} \right) dx\]
\[\int\limits_0^{2\pi} \cos^7 x dx\]
\[\int\limits_{\pi/6}^{\pi/2} \frac{\ cosec x \cot x}{1 + {cosec}^2 x} dx\]
\[\int\limits_0^2 \left( 2 x^2 + 3 \right) dx\]
Evaluate the following using properties of definite integral:
`int_(- pi/4)^(pi/4) x^3 cos^3 x "d"x`
Evaluate the following integrals as the limit of the sum:
`int_1^3 x "d"x`
Integrate `((2"a")/sqrt(x) - "b"/x^2 + 3"c"root(3)(x^2))` w.r.t. x
Given `int "e"^"x" (("x" - 1)/("x"^2)) "dx" = "e"^"x" "f"("x") + "c"`. Then f(x) satisfying the equation is:
