Advertisements
Advertisements
प्रश्न
Evaluate the following:
f(x) = `{{:("c"x",", 0 < x < 1),(0",", "otherwise"):}` Find 'c" if `int_0^1 "f"(x) "d"x` = 2
बेरीज
Advertisements
उत्तर
f(x) = `{{:("c"x",", 0 < x < 1),(0",", "otherwise"):}`
⇒ `int_0^1 "f"(x) "d"x` = 2
⇒ `int_0^2 "c"x "d"x` = 2
`"c"[x^2/2]_0^1` = 2
`"c"[1/2 - 0]` = 2
`1/2` = 2
⇒ c = 4
shaalaa.com
या प्रश्नात किंवा उत्तरात काही त्रुटी आहे का?
APPEARS IN
संबंधित प्रश्न
\[\int\limits_0^1 x \left( 1 - x \right)^5 dx\]
\[\int\limits_0^{2\pi} e^x \cos\left( \frac{\pi}{4} + \frac{x}{2} \right) dx\]
\[\int_0^\frac{\pi}{4} \left( \tan x + \cot x \right)^{- 2} dx\]
\[\int\limits_0^1 x e^{x^2} dx\]
\[\int\limits_0^{\pi/2} \frac{x + \sin x}{1 + \cos x} dx\]
\[\int\limits_0^\pi \frac{x \tan x}{\sec x \ cosec x} dx\]
\[\int\limits_{- 1}^1 \left( x + 3 \right) dx\]
\[\int\limits_0^{\pi/2} x \sin x\ dx\] is equal to
\[\int\limits_0^\infty \frac{x}{\left( 1 + x \right)\left( 1 + x^2 \right)} dx\]
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")`
