Advertisements
Advertisements
Question
If f'(x) = x2 + 5 and f(0) = −1, then find the value of f(x).
Advertisements
Solution
f'(x) = x2 + 5 ...(Given)
∴ f(x) = ∫f'(x) dx
∴ f(x) = ∫(x2 + 5) dx
∴ f(x) = ∫ x2 dx + 5 ∫ dx
∴ f(x) = `"x"^3/3 + 5"x" + "c"` ....(i)
Substitute x = 0, f(0) = −1 ...(Given)
∴ f(x) = `"x"^3/3 + 5"x" + "c"`
∴ f(0) = `0^3/3 + 5(0) + "c"`
∴ −1 = 0 + 0 + c
∴ c = −1
Substituting c = – 1 in (i), we get,
∴ f(x) = `"x"^3/3 + 5"x" + (− 1)`
∴ f(x) = `"x"^3/3 + 5"x" − 1`
APPEARS IN
RELATED QUESTIONS
Integrate the functions:
sin x ⋅ sin (cos x)
Integrate the functions:
sin (ax + b) cos (ax + b)
Integrate the functions:
`e^(2x+3)`
Integrate the functions:
tan2(2x – 3)
Write a value of
Write a value of\[\int\frac{\sin x + \cos x}{\sqrt{1 + \sin 2x}} dx\]
Write a value of\[\int\frac{\sin 2x}{a^2 \sin^2 x + b^2 \cos^2 x} \text{ dx }\]
Prove that: `int "dx"/(sqrt("x"^2 +"a"^2)) = log |"x" +sqrt("x"^2 +"a"^2) | + "c"`
Integrate the following w.r.t. x:
`3 sec^2x - 4/x + 1/(xsqrt(x)) - 7`
Evaluate the following integrals:
`int (cos2x)/sin^2x dx`
Evaluate the following integrals : `int sinx/(1 + sinx)dx`
Integrate the following functions w.r.t. x : `(e^(2x) + 1)/(e^(2x) - 1)`
Integrate the following functions w.r.t. x : `sin(x - a)/cos(x + b)`
Choose the correct options from the given alternatives :
`int dx/(cosxsqrt(sin^2x - cos^2x))*dx` =
Evaluate the following.
`int ("e"^"x" + "e"^(- "x"))^2 ("e"^"x" - "e"^(-"x"))`dx
Evaluate the following.
`int ("2x" + 6)/(sqrt("x"^2 + 6"x" + 3))` dx
`int cos sqrtx` dx = _____________
`int (sin4x)/(cos 2x) "d"x`
General solution of `(x + y)^2 ("d"y)/("d"x) = "a"^2, "a" ≠ 0` is ______. (c is arbitrary constant)
`int(sin2x)/(5sin^2x+3cos^2x) dx=` ______.
The general solution of the differential equation `(1 + y/x) + ("d"y)/(d"x)` = 0 is ______.
Find `int (x + 2)/sqrt(x^2 - 4x - 5) dx`.
Find : `int sqrt(x/(1 - x^3))dx; x ∈ (0, 1)`.
Evaluate the following
`int x^3/sqrt(1+x^4) dx`
Evaluate:
`int sin^3x cos^3x dx`
Evaluate `int 1/(x(x-1))dx`
Evaluate `int(1+x+x^2/(2!))dx`
Evaluate:
`intsqrt(sec x/2 - 1)dx`
For \[t=\cos x\], what is \[dt\]?
