Advertisements
Advertisements
प्रश्न
Evaluate the definite integral:
`int_0^(pi/4) (2 sec^2 x + x^3 + 2) dx`
Advertisements
उत्तर
Let `I int_0^(pi/4) (2 sec^2 x+ x^3 + 2) dx`
`[2 tan x x^4/4 + 2x]_0^(pi/4)`
`= 2 (tan pi/4 - tan0) + 1/4 (pi^4/256 - 0) + 2 (pi/4 - 0)`
`= 2 (1 - 0) + pi^4/1024 + pi/2`
`= pi^4/1024 + pi/2 + 2`
APPEARS IN
संबंधित प्रश्न
Evaluate : `∫_0^(π/2)(sin^2 x)/(sinx+cosx)dx`
Evaluate the definite integral:
`int_(-1)^1 (x + 1)dx`
Evaluate the definite integral:
`int_2^3 1/x dx`
Evaluate the definite integral:
`int_1^2 (4x^3 - 5x^2 + 6x + 9) dx`
Evaluate the definite integral:
`int_0^(pi/2) cos 2x dx`
Evaluate the definite integral:
`int_4^5 e^x dx`
Evaluate the definite integral:
`int_0^(pi/4) tan x dx`
Evaluate the definite integral:
`int_0^1 dx/(1+x^2)`
Evaluate the definite integral:
`int_0^(pi/2) cos^2 xdx`
Evaluate the definite integral:
`int_2^3 (xdx)/(x^2 + 1)`
Evaluate the definite integral:
`int_0^1 (2x + 3)/(5x^2 + 1) dx`
Evaluate the definite integral:
`int_0^1 x e^(x^2) dx`
Evaluate the definite integral:
`int_0^pi (sin^2 x/2 - cos^2 x/2) dx`
Evaluate the definite integral:
`int_0^2 (6x +3)/(x^2 + 4)` dx
`int_1^(sqrt3)dx/(1+x^2) ` equals:
`int_0^(2/3) dx/(4+9x^2)` equals:
Evaluate : \[\int\frac{x \cos^{- 1} x}{\sqrt{1 - x^2}}dx\] .
Evaluate:
`int_0^π(sin^4x + cos^4x)dx`
Hence evaluate:
`int_(-2π)^(2π) (sin^4x + cos^4x)/(1 + e^x)dx`
If \(f\) is continuous on \([a,b]\) and \[A(x)=\int_a^x f(t)\,dt,\] what is \(A'(x)\) for every \(x\) in \((a,b)\)?
Why is there no need to write the constant of integration \(C\) while evaluating a definite integral?
Which expression correctly applies the limits to a definite integral?
For \[\int_4^9\frac{\sqrt{x}}{(30-x^{\frac32})^2}\,dx,\] which substitution is used?
An antiderivative of \[\frac{\sqrt{x}}{(30-x^{\frac32})^2}\] is:
Evaluate \[\int_4^9\frac{\sqrt{x}}{(30-x^{\frac32})^2}\,dx.\]
For \[\int_0^{\frac\pi4}\sin^3 2t\cos 2t\,dt,\] which substitution and differential relation are correct?
Which function is an antiderivative of \[\sin^3 2t\cos 2t?\]
Evaluate \[\int_0^{\frac\pi4}\sin^3 2t\cos 2t\,dt.\]
