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
योग
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
Definite Integrals
क्या इस प्रश्न या उत्तर में कोई त्रुटि है?
APPEARS IN
संबंधित प्रश्न
\[\int_0^\frac{\pi}{4} \left( \tan x + \cot x \right)^{- 2} dx\]
\[\int\limits_0^5 \frac{\sqrt[4]{x + 4}}{\sqrt[4]{x + 4} + \sqrt[4]{9 - x}} dx\]
\[\int\limits_0^{\pi/2} \frac{1}{1 + \sqrt{\tan x}} dx\]
\[\int\limits_0^\pi \frac{x}{1 + \cos \alpha \sin x} dx, 0 < \alpha < \pi\]
\[\int\limits_{- \pi/4}^{\pi/4} \sin^2 x\ dx\]
\[\int\limits_0^1 \sqrt{x \left( 1 - x \right)} dx\] equals
\[\int\limits_1^5 \frac{x}{\sqrt{2x - 1}} dx\]
\[\int\limits_0^1 \frac{1 - x}{1 + x} dx\]
\[\int\limits_1^2 \frac{x + 3}{x\left( x + 2 \right)} dx\]
Evaluate `int (x^2"d"x)/(x^4 + x^2 - 2)`
