Advertisements
Advertisements
Question
If f '(x) = `1/"x" + "x"` and f(1) = `5/2`, then f(x) = log x + `"x"^2/2` + ______
Advertisements
Solution
If f '(x) = `1/"x" + "x"` and f(1) = `5/2`, then f(x) = log x + `"x"^2/2` + 2
Explanation:
f(x) = ∫f '(x) dx
`= int (1/"x" + "x")` dx
f(x) = log |x| + `"x"^2/2 + "c"` ...(i)
f(1) = `5/2`
f(1) = log 1 + `1^2/2` + c
∴ `5/2 = 0 + 1/2 + "c"` ...(∵ log 1 = 0)
∴ c = `5/2 - 1/2`
∴ = `4/2` = 2
∴ c = 2
∴ f(x) = log |x| + `"x"^2/2` + 2
Notes
The answer in the textbook is incorrect.
RELATED QUESTIONS
Evaluate : `int (sinx)/sqrt(36-cos^2x)dx`
Evaluate :`intxlogxdx`
Evaluate: `int sqrt(tanx)/(sinxcosx) dx`
Integrate the functions:
`(e^(2x) - 1)/(e^(2x) + 1)`
Solve:
dy/dx = cos(x + y)
Write a value of
Write a value of\[\int\text{ tan x }\sec^3 x\ dx\]
Evaluate the following : `int (1)/sqrt(3x^2 + 5x + 7).dx`
Choose the correct options from the given alternatives :
`int (e^x(x - 1))/x^2*dx` =
Choose the correct options from the given alternatives :
`int f x^x (1 + log x)*dx`
If f'(x) = 4x3 − 3x2 + 2x + k, f(0) = 1 and f(1) = 4, find f(x).
Evaluate the following.
`int "x" sqrt(1 + "x"^2)` dx
Choose the correct alternative from the following.
The value of `int "dx"/sqrt"1 - x"` is
Fill in the Blank.
`int (5("x"^6 + 1))/("x"^2 + 1)` dx = x4 + ______ x3 + 5x + c
Evaluate: `int 1/(2"x" + 3"x" log"x")` dx
Evaluate: `int log ("x"^2 + "x")` dx
`int x^2/sqrt(1 - x^6)` dx = ________________
`int 2/(sqrtx - sqrt(x + 3))` dx = ________________
If f'(x) = `x + 1/x`, then f(x) is ______.
If `int(cosx - sinx)/sqrt(8 - sin2x)dx = asin^-1((sinx + cosx)/b) + c`. where c is a constant of integration, then the ordered pair (a, b) is equal to ______.
The value of `intsinx/(sinx - cosx)dx` equals ______.
Evaluate `int(1 + x + x^2/(2!) )dx`
Evaluate the following.
`int x^3/(sqrt(1+x^4))dx`
Evaluate the following.
`int(20 - 12"e"^"x")/(3"e"^"x" - 4) "dx"`
Evaluate the following.
`intxsqrt(1+x^2)dx`
Evaluate `int(1+x+x^2/(2!))dx`
Evaluate the following.
`int1/(x^2 + 4x-5)dx`
For \[\sqrt{\frac{x-\alpha}{\beta-x}}\] or \[\sqrt{(x-\alpha)(\beta-x)}\], where \[\beta>\alpha\], which substitution is used?
