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Integral Calculus

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Question : 58 of 73

Marks: +1, -0

The equation of the curve which passes through the point (1,3) and has the slope

4 x-3

(x y)

$y=2 x^{3}-3 x+4$
$y=2 x^{2}-3 x+4$
$x=2 y^{2}-3 y+4$
None of these

Solution:

Concept:

Integrate the slope to get the curve equation, then use the given point to find the constant.

Explanation:

Slope is

\frac{dy}{dx} = 4x - 3

.

Integrate:

y = \int (4x - 3) \, dx = 2x^2 - 3x + C

.

Curve passes through

(1,3)

:

3 = 2(1)^2 - 3(1) + C

3 = 2 - 3 + C \Rightarrow C = 4

.

Hence

y = 2x^2 - 3x + 4

.

Answer:

y = 2x^2 - 3x + 4

corresponds to option B.

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