Advertisements
Advertisements
प्रश्न
Advertisements
उत्तर
\[\int_0^3 \frac{1}{x^2 + 9} d x\]
\[ = \int_0^3 \frac{1}{x^2 + 3^2} d x\]
\[ = \frac{1}{3} \left[ \tan^{- 1} \frac{x}{3} \right]_0^3 \]
\[ = \frac{1}{3}\left( \tan^{- 1} 1 - \tan^{- 1} 0 \right)\]
\[ = \frac{1}{3}\left( \frac{\pi}{4} - 0 \right)\]
\[ = \frac{\pi}{12}\]
APPEARS IN
संबंधित प्रश्न
Evaluate each of the following integral:
The value of \[\int\limits_0^{\pi/2} \cos x\ e^{\sin x}\ dx\] is
If \[\int\limits_0^a \frac{1}{1 + 4 x^2} dx = \frac{\pi}{8},\] then a equals
The value of the integral \[\int\limits_{- 2}^2 \left| 1 - x^2 \right| dx\] is ________ .
The value of \[\int\limits_0^\pi \frac{1}{5 + 3 \cos x} dx\] is
`int_0^(2a)f(x)dx`
\[\int\limits_0^1 \tan^{- 1} \left( \frac{2x}{1 - x^2} \right) dx\]
\[\int\limits_1^2 \frac{x + 3}{x\left( x + 2 \right)} dx\]
\[\int\limits_0^{\pi/4} e^x \sin x dx\]
\[\int\limits_{- \pi}^\pi x^{10} \sin^7 x dx\]
\[\int\limits_{- 1}^1 e^{2x} dx\]
Prove that `int_a^b ƒ ("x") d"x" = int_a^bƒ(a + b - "x") d"x" and "hence evaluate" int_(π/6)^(π/3) (d"x")/(1+sqrt(tan "x")`
Evaluate the following:
f(x) = `{{:("c"x",", 0 < x < 1),(0",", "otherwise"):}` Find 'c" if `int_0^1 "f"(x) "d"x` = 2
Evaluate the following using properties of definite integral:
`int_(- pi/4)^(pi/4) x^3 cos^3 x "d"x`
Evaluate the following:
`int_0^oo "e"^(- x/2) x^5 "d"x`
Choose the correct alternative:
If f(x) is a continuous function and a < c < b, then `int_"a"^"c" f(x) "d"x + int_"c"^"b" f(x) "d"x` is
Choose the correct alternative:
If n > 0, then Γ(n) is
If `int (3"e"^x - 5"e"^-x)/(4"e"6x + 5"e"^-x)"d"x` = ax + b log |4ex + 5e –x| + C, then ______.
