Advertisements
Advertisements
प्रश्न
\[\int\limits_0^1 \log\left( 1 + x \right) dx\]
Advertisements
उत्तर
\[\int_0^1 \log\left( 1 + x \right) d x\]
\[ = \int_0^1 \log\left( 1 + x \right) \times 1 d x\]
\[ = \left[ \log\left( 1 + x \right) x \right]_0^1 - \int_0^1 \frac{x}{1 + x}dx\]
\[ = \left[ \log\left( 1 + x \right) x \right]_0^1 - \int_0^1 \left( 1 - \frac{1}{1 + x} \right)dx\]
\[ = \left[ x\log\left( 1 + x \right) \right]_0^1 - \left[ x - \log\left( 1 + x \right) \right]_0^1 \]
\[ = \log2 - 1 + \log2\]
\[ = 2\log2 - 1\]
\[ = \log4 - \log e\]
\[ = \log\frac{4}{e}\]
APPEARS IN
संबंधित प्रश्न
Evaluate the following integral:
Evaluate each of the following integral:
Solve each of the following integral:
\[\int\limits_0^\infty \frac{1}{1 + e^x} dx\] equals
The derivative of \[f\left( x \right) = \int\limits_{x^2}^{x^3} \frac{1}{\log_e t} dt, \left( x > 0 \right),\] is
If \[I_{10} = \int\limits_0^{\pi/2} x^{10} \sin x\ dx,\] then the value of I10 + 90I8 is
\[\int\limits_0^{2a} f\left( x \right) dx\] is equal to
\[\int\limits_0^{\pi/4} \tan^4 x dx\]
\[\int\limits_0^a \frac{\sqrt{x}}{\sqrt{x} + \sqrt{a - x}} dx\]
\[\int\limits_2^3 \frac{\sqrt{x}}{\sqrt{5 - x} + \sqrt{x}} dx\]
\[\int\limits_0^\pi \frac{dx}{6 - \cos x}dx\]
Using second fundamental theorem, evaluate the following:
`int_1^"e" ("d"x)/(x(1 + logx)^3`
Evaluate the following:
`int_0^2 "f"(x) "d"x` where f(x) = `{{:(3 - 2x - x^2",", x ≤ 1),(x^2 + 2x - 3",", 1 < x ≤ 2):}`
Evaluate the following:
`int_(-1)^1 "f"(x) "d"x` where f(x) = `{{:(x",", x ≥ 0),(-x",", x < 0):}`
Evaluate the following:
f(x) = `{{:("c"x",", 0 < x < 1),(0",", "otherwise"):}` Find 'c" if `int_0^1 "f"(x) "d"x` = 2
If f(x) = `{{:(x^2"e"^(-2x)",", x ≥ 0),(0",", "otherwise"):}`, then evaluate `int_0^oo "f"(x) "d"x`
Choose the correct alternative:
`int_0^1 (2x + 1) "d"x` is
Find `int sqrt(10 - 4x + 4x^2) "d"x`
Given `int "e"^"x" (("x" - 1)/("x"^2)) "dx" = "e"^"x" "f"("x") + "c"`. Then f(x) satisfying the equation is:
Find: `int logx/(1 + log x)^2 dx`
