How do you find y from \frac{dy}{dx}?

Integrate: y=\int\frac{dy}{dx}\,dx, including +c.

Why does the gradient function alone not fix the curve?

Any vertical translation of the curve has the same gradient function; the constant c is undetermined.

Substitute the coordinates of a point on the curve into the integrated equation.

How can you check your equation of the curve?

Differentiate it to get the gradient function, and check the given point satisfies it.

Two curves have the same gradient function. How do they differ?

Only in the constant c: one is a vertical translation of the other.

Where do you substitute the point: into \frac{dy}{dx} or y?

Into the equation for y (the integrated one with c).

Finding a curve from its gradientEdexcel International A Level Maths: Flashcards

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How do you find $y$ from $\frac{\mathrm{d}y}{\mathrm{d}x}$?

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How do you find $y$ from $\frac{\mathrm{d}y}{\mathrm{d}x}$ ?: Integrate: $y=\int\frac{\mathrm{d}y}{\mathrm{d}x}\,\mathrm{d}x$ , including $+c$ .
Why does the gradient function alone not fix the curve?: Any vertical translation of the curve has the same gradient function; the constant $c$ is undetermined.
How do you find $c$ ?: Substitute the coordinates of a point on the curve into the integrated equation.
Integrate $3x^2-4x+1$ .: $x^3-2x^2+x+c$
Curve through $(2,9)$ with $\frac{\mathrm{d}y}{\mathrm{d}x}=3x^2-4x+1$ : $c$ ?: $c=7$
Integrate $\frac{8}{x^3}$ .: $-4x^{-2}+c$
Integrate $-3\sqrt{x}$ .: $-2x^{\frac32}+c$
What should you do first if the gradient is $(2x+3)(x-1)$ ?: Expand it to $2x^2+x-3$ .
How can you check your equation of the curve?: Differentiate it to get the gradient function, and check the given point satisfies it.
Two curves have the same gradient function. How do they differ?: Only in the constant $c$ : one is a vertical translation of the other.
Where do you substitute the point: into $\frac{\mathrm{d}y}{\mathrm{d}x}$ or $y$ ?: Into the equation for $y$ (the integrated one with $c$ ).
Find $y$ if $\frac{\mathrm{d}y}{\mathrm{d}x}=4x-1$ and $y=3$ when $x=0$ .: $y=2x^2-x+3$
What is $4^{\frac32}$ ?: $8$

Exam questions on Finding a curve from its gradient

A curve passes through the point $(2,9)$ and has gradient function $\dfrac{\mathrm{d}y}{\mathrm{d}x}=3x^2-4x+1$ .
Find the value of $y$ on the curve when $x=-1$ .2 marks
The gradient of a curve at the point $(x,y)$ is given by $\dfrac{\mathrm{d}y}{\mathrm{d}x}=\dfrac{8}{x^3}-3\sqrt{x}$ for $x>0$ . The curve passes through the point $(1,4)$ .
Find the value of $y$ on the curve when $x=4$ .2 marks
The curve $y=f(x)$ has gradient function $f'(x)=(2x+3)(x-1)$ and passes through the point $(3,10)$ .
Show that $f(x)=\frac23x^3+\frac12x^2-3x+c$ , where $c$ is a constant.3 marks

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Written by the Exaim team, led by Shaun Daswani (Head of Upper Secondary, Improve ME Institute; MSc Financial Mathematics, Imperial College London; BSc, UCL) and Jason Daswani (operational lead, Improve ME Institute; LSE).