Advertisements
Advertisements
प्रश्न
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
If `f` is an integrable function such that f(2a − x) = f(x), then prove that
If f(2a − x) = −f(x), prove that
If f is an integrable function, show that
\[\int\limits_{- a}^a f\left( x^2 \right) dx = 2 \int\limits_0^a f\left( x^2 \right) dx\]
The value of the integral \[\int\limits_0^\infty \frac{x}{\left( 1 + x \right)\left( 1 + x^2 \right)} dx\]
The derivative of \[f\left( x \right) = \int\limits_{x^2}^{x^3} \frac{1}{\log_e t} dt, \left( x > 0 \right),\] is
The value of \[\int\limits_0^\pi \frac{1}{5 + 3 \cos x} dx\] is
The value of \[\int\limits_0^1 \tan^{- 1} \left( \frac{2x - 1}{1 + x - x^2} \right) dx,\] is
\[\int\limits_0^1 \cos^{- 1} x dx\]
\[\int\limits_1^2 \frac{1}{x^2} e^{- 1/x} dx\]
\[\int\limits_0^1 \log\left( 1 + x \right) dx\]
\[\int\limits_0^{2\pi} \cos^7 x dx\]
\[\int\limits_1^4 \left( x^2 + x \right) dx\]
Using second fundamental theorem, evaluate the following:
`int_1^2 (x "d"x)/(x^2 + 1)`
Using second fundamental theorem, evaluate the following:
`int_1^2 (x - 1)/x^2 "d"x`
Choose the correct alternative:
The value of `int_(- pi/2)^(pi/2) cos x "d"x` is
Choose the correct alternative:
Γ(1) 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 ______.
If x = `int_0^y "dt"/sqrt(1 + 9"t"^2)` and `("d"^2y)/("d"x^2)` = ay, then a equal to ______.
