Animated Solution for Mathematics - Definite Integration: The area of the region enclosed by the curve $y=x^3$ and its tangent at the point $(-1,-1)$ is

A JEE Masterclass

Welcome, future engineer. Today, we are not just solving a problem; we are witnessing a beautiful interaction between a cubic curve and a linear tangent. In the world of JEE Advanced, problems like this are not just about calculation—they are about understanding the 'soul' of the functions involved.

Let us embark on this journey to find the area enclosed by $y=x^3$ and its tangent at $P(-1,-1)$.

The Calculus of Tangency

Imagine the curve $y=x^3$. It is elegant, symmetric, and smooth. We are anchored at the point $P(-1,-1)$.

To find the tangent line, we need to know how the curve is behaving at that exact moment. This is where the derivative, our most powerful tool, comes into play. We differentiate the function $y=x^3$ with respect to $x$ to find the slope generator:

This expression, $3x^2$, tells us the slope of the tangent at any point $x$. But we are interested in the specific moment when $x=-1$. Substituting this into our slope generator, we get:

With a slope of $3$ and a point $(-1,-1)$, we invoke the point-slope form of a linear equation: $y-y_1=m(x-x_1)$. Substituting our values:

There it is. The tangent line is $y=3x+2$. It is a straight line cutting through the coordinate plane, destined to intersect our cubic curve once more.

The Algebraic Hunt

Now, we must find where this line meets the curve again. This is the 'intersection' phase. We set the two equations equal to each other:

Rearranging this gives us a cubic equation:

Here is where the JEE aspirant separates themselves from the crowd. You could perform long division, but you already know a secret: the line is tangent at $x=-1$. This means the line touches the curve, which implies that $x=-1$ is a repeated root. Therefore, $(x+1{)}^2$ must be a factor of our cubic polynomial.

By dividing $x^3-3x-2$ by $(x+1{)}^2$, we quickly find the remaining factor is $(x-2)$. Thus, our equation becomes:

This reveals our second intersection point: $x=2$. The tangent line and the curve embrace at $x=-1$ and part ways at $x=2$. We have our limits of integration!

The Summation of Slices

We are now ready to calculate the area. We are looking for the area between the line and the curve from $x=-1$ to $x=2$. As we discussed, the line $y=3x+2$ sits above the curve $y=x^3$ in this interval.

The area $A$ is the integral of the difference:

Let us integrate term by term. This is the moment of truth where precision matters most:

Now, we evaluate at the upper limit ($x=2$):

Next, we evaluate at the lower limit ($x=-1$):

To subtract these, we find a common denominator of $4$:

Finally, we subtract the lower limit result from the upper limit result:

The Elegance of the Result

The final answer is $\frac{27}{4}$ square units.

Look at that number. It is not just a fraction; it is the culmination of understanding tangency, polynomial roots, and the power of definite integration. You didn't just calculate an area; you mapped the relationship between two functions. Keep this mindset—always look for the geometric meaning behind the algebraic steps. You are ready for the next challenge.