If $\frac{d}{dx}[f(x)] = ax + b$ and $f(0) = 0$, then $f(x)$ is equal to: (a) $a + b$ (b) $\frac{ax^2}{2} + bx$ (c) $\frac{ax^2}{2} + bx + c$ (d) $b$

If (d)/(dx)[f(x)] = ax + b and f(0) = 0, then f(x) is equal to: (a) a + b (b) (ax²)/(2) + bx (c) (ax²)/(2) + b

Question

\frac{d}{dx}[f(x)] = ax + b

(a)

a + b

(b)

\frac{ax^2}{2} + bx

(c)

\frac{ax^2}{2} + bx + c

(d)

b

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Integrate the derivative and use the initial condition $f(0) = 0$ to determine the constant; $f(x) = \frac{ax^2}{2} + bx$ .

When you know the derivative of a function, you can recover the original function through integration. The process introduces an arbitrary constant of integration, which you then pin down using any given initial or boundary condition.

Here we're told that $\frac{d}{dx}[f(x)] = ax + b$ , which means the rate of change of $f$ is a linear function of $x$ . To find $f(x)$ itself, we integrate both sides with respect to $x$ .

Finding $f(x)$ by integration

Integrate the derivative.

f(x) = \int (ax + b) \, dx

Applying the power rule term by term:

f(x) = a \cdot \frac{x^2}{2} + bx + C

where $C$ is the constant of integration that appears whenever we perform an indefinite integral.

Apply the initial condition $f(0) = 0$ .

Substitute $x = 0$ into the expression we just found:

f(0) = a \cdot \frac{0^2}{2} + b \cdot 0 + C = C

Since we're given that $f(0) = 0$ , we have: …

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Same / Similar Concept — real previous-year questions on the same or a closely similar concept, not this exact question.

CBSE 2025Set ANNUAL1 markMCQ
$\dfrac{dy}{dx}=\cos x$
(i) $y=\sin x-2$
(ii) $y=\sin x+1$
(iii) $y=\sin x+2$
(iv) $y=\sin x-1$
›Reveal solutionSolution

Integrate directly and use the initial condition to find $C$ .

$\frac{dy}{dx} = \cos x \implies y = \int \cos x\,dx = \sin x + C$

Apply the initial condition $x=0,\ y=2$ : …

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CBSE 2025Set ANNUAL1 markMCQ
$\dfrac{dy}{dx}+x=0$
(a) $y+\dfrac{x^2}{2}+1=0$
(b) $y+\dfrac{x^2}{2}=1$
(c) $y+2x^2+1=0$
(d) $y+2x^2=1$
›Reveal solutionSolution

Separate variables, integrate, and use the initial condition $x=0,y=1$ to fix the constant.

$\dfrac{dy}{dx}=-x \;\Rightarrow\; dy=-x\,dx \;\Rightarrow\; \int dy=-\int x\,dx$

$y=-\dfrac{x^2}{2}+C$

At $x=0,y=1$ : $1=0+C\Rightarrow C=1$ .

…

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CBSE 2024Set ANNUAL1 markMCQ
$\dfrac{d}{dx}f(x) = 4x^3 - \dfrac{3}{x^4}$
(a) $x^4 + \dfrac{1}{x^3} - \dfrac{129}{8}$
(b) $x^3 + \dfrac{1}{x^4} + \dfrac{129}{8}$
(c) $x^4 - \dfrac{1}{x^3} + \dfrac{129}{8}$
(d) $x^3 + \dfrac{1}{x^4} - \dfrac{129}{8}$
›Reveal solutionSolution

Integrate $f'(x)$ term by term, then use $f(2)=0$ to find the constant of integration.

Given $\dfrac{d}{dx}f(x) = 4x^3 - \dfrac{3}{x^4} = 4x^3 - 3x^{-4}$ .

Integrating:

$f(x) = \int (4x^3 - 3x^{-4})\,dx = x^4 - 3\cdot\frac{x^{-3}}{-3} + C = x^4 + x^{-3} + C = x^4 + \frac{1}{x^3} + C$

Using $f(2)=0$ : …

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CBSE 2023Set 65/2/11 markMCQ
$\frac{d}{dx}[f(x)] = ax + b$
(a) $a + b$
(b) $\frac{ax^2}{2} + bx$
(c) $\frac{ax^2}{2} + bx + c$
(d) $b$
›Reveal solutionSolution
Integrate the derivative and use the initial condition $f(0) = 0$ to determine the constant; $f(x) = \frac{ax^2}{2} + bx$ .

When you know the derivative of a function, you can recover the original function through integration. The process introduces an arbitrary constant of integration, which you then pin down using any given initial or boundary condition.

Here we're told that $\frac{d}{dx}[f(x)] = ax + b$ , which means the rate of change of $f$ is a linear function of $x$ . To find $f(x)$ itself, we integrate both sides with respect to $x$ .

Finding $f(x)$ by integration

Integrate the derivative.

$f(x) = \int (ax + b) \, dx$
Applying the power rule term by term:
$f(x) = a \cdot \frac{x^2}{2} + bx + C$
where $C$ is the constant of integration that appears whenever we perform an indefinite integral.

Apply the initial condition $f(0) = 0$ .

Substitute $x = 0$ into the expression we just found:

$f(0) = a \cdot \frac{0^2}{2} + b \cdot 0 + C = C$
Since we're given that $f(0) = 0$ , we have: …
Start your 14-day free trial to see the full solution →

Q.If $\frac{d}{dx}[f(x)] = ax + b$ and $f(0) = 0$ , then $f(x)$ is equal to:

Concept understanding — Initial Value Problem

Initial Value Problem

The intuition: a rule of change plus a starting point

The precise statement

A worked example

Finding $f(x)$ by integration

Finding $f(x)$ by integration

Q.If ddx[f(x)]=ax+b\frac{d}{dx}[f(x)] = ax + bdxd​[f(x)]=ax+b and f(0)=0f(0) = 0f(0)=0, then f(x)f(x)f(x) is equal to:

Finding f(x)f(x)f(x) by integration

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Q.If $\frac{d}{dx}[f(x)] = ax + b$ and $f(0) = 0$ , then $f(x)$ is equal to:

Finding $f(x)$ by integration

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