Area of region enclosed by curve y=x3 and its tangent at (–1,–1)
Area of region enclosed by curve y=x3 and its tangent at (–1,–1)
4
27
\(\frac{4}{27}\)
\(\frac{27}{4}\)
The Correct Option is D
Solution and Explanation
The correct answer is option (D): \(\frac{27}{4}\)
y+1=3(x+1)
i.e y = 3x+2
Point of intersection with curve (2,8)
So area = \(\int_{-1}^{2}((3x+2)-x^3)dx=\frac{27}{4}\)
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Concepts Used:
Application of Derivatives
Various Applications of Derivatives-
Rate of Change of Quantities:
If some other quantity ‘y’ causes some change in a quantity of surely ‘x’, in view of the fact that an equation of the form y = f(x) gets consistently pleased, i.e, ‘y’ is a function of ‘x’ then the rate of change of ‘y’ related to ‘x’ is to be given by
\(\frac{\triangle y}{\triangle x}=\frac{y_2-y_1}{x_2-x_1}\)
This is also known to be as the Average Rate of Change.
Increasing and Decreasing Function:
Consider y = f(x) be a differentiable function (whose derivative exists at all points in the domain) in an interval x = (a,b).
- If for any two points x1 and x2 in the interval x such a manner that x1 < x2, there holds an inequality f(x1) ≤ f(x2); then the function f(x) is known as increasing in this interval.
- Likewise, if for any two points x1 and x2 in the interval x such a manner that x1 < x2, there holds an inequality f(x1) ≥ f(x2); then the function f(x) is known as decreasing in this interval.
- The functions are commonly known as strictly increasing or decreasing functions, given the inequalities are strict: f(x1) < f(x2) for strictly increasing and f(x1) > f(x2) for strictly decreasing.
Read More: Application of Derivatives