Advertisements
Advertisements
प्रश्न
Advertisements
उत्तर
\[Let\ I = \int_0^7 \frac{\sqrt[3]{x}}{\sqrt[3]{x} + \sqrt[3]{7 - x}} d x ..................(1)\]
\[ = \int_0^7 \frac{\sqrt[3]{7 - x}}{\sqrt[3]{7 - x} + \sqrt[3]{x}} dx .................\left(\text{Using }\int_0^a f\left( x \right) dx = \int_0^a f\left( a - x \right) dx \right)\]
\[ = \int_0^7 \frac{\sqrt[3]{7 - x}}{\sqrt[3]{x} + \sqrt[3]{7 - x}} dx ..................(2)\]
\[\text{Adding (1) and (2) we get}\]
\[2I = \int_0^7 \frac{\sqrt[3]{x} + \sqrt[3]{7 - x}}{\sqrt[3]{x} + \sqrt[3]{7 - x}} d x \]
\[ = \int_0^7 dx\]
\[ = \left[ x \right]_0^7 = 7\]
\[Hence\ I = \frac{7}{2}\]
APPEARS IN
संबंधित प्रश्न
The value of \[\int\limits_0^\pi \frac{x \tan x}{\sec x + \cos x} dx\] is __________ .
The value of the integral \[\int\limits_0^{\pi/2} \frac{\sqrt{\cos x}}{\sqrt{\cos x} + \sqrt{\sin x}} dx\] is
`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
\[\int\limits_0^1 x \left( \tan^{- 1} x \right)^2 dx\]
\[\int\limits_0^\pi \frac{x}{a^2 - \cos^2 x} dx, a > 1\]
\[\int\limits_0^4 x dx\]
Find : `∫_a^b logx/x` dx
Evaluate the following using properties of definite integral:
`int_0^1 log (1/x - 1) "d"x`
Evaluate the following:
`int_0^oo "e"^(-4x) x^4 "d"x`
Choose the correct alternative:
`int_0^1 (2x + 1) "d"x` is
Choose the correct alternative:
If n > 0, then Γ(n) is
Choose the correct alternative:
`Γ(3/2)`
Choose the correct alternative:
`int_0^oo x^4"e"^-x "d"x` is
`int x^9/(4x^2 + 1)^6 "d"x` is equal to ______.
Evaluate: `int_(-1)^2 |x^3 - 3x^2 + 2x|dx`
