Advertisements
Advertisements
प्रश्न
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
विकल्प
`int_"a"^"b" f(x) "d"x - int_"a"^"c" f(x) "d"x`
`int_"a"^"c" f(x) "d"x - int_"a"^"b" f(x) "d"x`
`int_"a"^"b" f(x) "d"x`
0
MCQ
Advertisements
उत्तर
`int_"a"^"b" f(x) "d"x`
shaalaa.com
Definite Integrals
क्या इस प्रश्न या उत्तर में कोई त्रुटि है?
APPEARS IN
संबंधित प्रश्न
\[\int\limits_0^1 \frac{1}{2 x^2 + x + 1} dx\]
\[\int\limits_0^2 \frac{1}{\sqrt{3 + 2x - x^2}} dx\]
\[\int\limits_0^{\pi/2} \sin^3 x\ dx\]
\[\int_\frac{\pi}{6}^\frac{\pi}{3} \left( \tan x + \cot x \right)^2 dx\]
\[\int\limits_0^9 f\left( x \right) dx, where f\left( x \right) \begin{cases}\sin x & , & 0 \leq x \leq \pi/2 \\ 1 & , & \pi/2 \leq x \leq 3 \\ e^{x - 3} & , & 3 \leq x \leq 9\end{cases}\]
\[\int\limits_0^2 x\left[ x \right] dx .\]
\[\int\limits_0^{2a} f\left( x \right) dx\] is equal to
\[\int\limits_1^5 \frac{x}{\sqrt{2x - 1}} dx\]
\[\int\limits_0^{2\pi} \cos^7 x dx\]
\[\int\limits_0^{\pi/2} \frac{1}{2 \cos x + 4 \sin x} dx\]
