Let a and b be real constants such that the function f defined by f(x) = {x² + 3x + a , x ≤ 1; bx +

Q: Wait, why do we need to check for continuity if the problem already says it is differentiable?

That is a fantastic question! In calculus, differentiability is a 'stronger' condition than continuity. By definition, if a function is differentiable at a point, it *must* be continuous there. However, the reverse isn't always true. We check continuity because it provides us with a crucial algebraic equation involving $a$ and $b$. Without that 'glue' condition, we wouldn't have enough equations to solve for both variables.

Q: Why is the derivative of $bx+2$ just $b$?

Remember that $b$ is a constant. When you differentiate a linear term like $bx$ with respect to $x$, the $x$ becomes 1, leaving just the coefficient $b$. The constant term $2$ disappears because its rate of change is zero. So, the slope of that linear segment is constant, which is exactly $b$.

Q: Why do we have to split the integral at $x=1$?

We split the integral because the function changes its 'rule' at $x=1$. The area under the curve is the sum of the areas of two different geometric shapes: a parabola on the left and a line on the right. You cannot integrate a piecewise function in one go; you must integrate each piece over its specific domain.

Question

Let $a$ and $b$ be real constants such that the function $f$ defined by $f(x) = \begin{cases} x^2 + 3x + a & , x \le 1 \ bx + 2 & , x > 1 \end{cases}$ be differentiable on $\mathbb{R}$. Then, the value of $\int_{-2}^{2} f(x) dx$ equals

Options:

(A) $\frac{15}{6}$

(B) $\frac{19}{6}$

(C) 21

(D) 17

Sigma Percentile · Accepted Answer

### The Art of the Smooth Transition

Imagine you are an engineer designing a roller coaster track. You have two sections of track: a parabolic curve that brings the coaster down, and a straight, linear path that carries it to the station.

If these two tracks do not meet perfectly at the junction, the coaster would derail. In mathematics, we call this **continuity**.

Even if they meet, if there is a sharp "kink" or a sudden change in angle at the junction, the coaster would experience a violent jolt. To ensure a smooth ride, we need **differentiability**.

In this problem, we are given a piecewise function $f(x)$ that is differentiable everywhere. This tells us that at the junction point, $x=1$, the function is not only connected but also smooth. Let us uncover the constants $a$ and $b$ that make this possible.

### Phase 1: The Conditions of Smoothness

We are given the function:

$$f(x) = \begin{cases} x^2 + 3x + a & , x \le 1 \ bx + 2 & , x > 1 \end{cases}$$

To ensure the function is differentiable at $x=1$, we must satisfy the "glue" condition: the left-hand limit must equal the right-hand limit.

Approaching $x=1$ from the left gives $1^2 + 3(1) + a = 4 + a$. Approaching from the right gives $b(1) + 2 = b + 2$. For continuity, these must be equal:

$$4 + a = b + 2 \implies a - b = -2$$

### Phase 2: The Derivative Hunt

We have two unknowns, so we must invoke the "smoothness" condition. The derivative of the left piece must equal the derivative of the right piece at $x=1$.

For $x < 1$, the derivative of $x^2 + 3x + a$ is $2x + 3$. For $x > 1$, the derivative of $bx + 2$ is simply $b$. At the critical point $x=1$, these slopes must be identical:

$$2(1) + 3 = b \implies b = 5$$

With $b=5$ in hand, we return to our continuity equation: $a - 5 = -2$. Solving for $a$, we find $a = 3$.

The function is now fully defined as:

$$f(x) = \begin{cases} x^2 + 3x + 3 & , x \le 1 \ 5x + 2 & , x > 1 \end{cases}$$

### Phase 3: The Integration Journey

We are tasked with finding the area under the curve from $x=-2$ to $x=2$. Because the function changes its identity at $x=1$, we split our integral into two parts:

$$\int_{-2}^{2} f(x) dx = \int_{-2}^{1} (x^2 + 3x + 3) dx + \int_{1}^{2} (5x + 2) dx$$

First, we evaluate the parabolic section:

$$\int_{-2}^{1} (x^2 + 3x + 3) dx = \left[ \frac{x^3}{3} + \frac{3x^2}{2} + 3x ight]_{-2}^{1}$$

Evaluating at the limits:

$$\left( \frac{1}{3} + \frac{3}{2} + 3 ight) - \left( \frac{-8}{3} + \frac{12}{2} - 6 ight) = \frac{29}{6} - \left( -\frac{8}{3} ight) = \frac{29}{6} + \frac{16}{6} = \frac{45}{6} = 7.5$$

Next, we evaluate the linear section:

$$\int_{1}^{2} (5x + 2) dx = \left[ \frac{5x^2}{2} + 2x ight]_{1}^{2}$$

Evaluating at the limits:

$$\left( \frac{5(4)}{2} + 2(2) ight) - \left( \frac{5}{2} + 2 ight) = 14 - 4.5 = 9.5$$

### The Grand Finale

Finally, we sum these two areas to find the total integral:

$$7.5 + 9.5 = 17$$

We have navigated the constraints of differentiability, solved for our constants, and calculated the area under the curve. The final result is **17**.

Animated Solution for Mathematics - Limits, Continuity and Differentiability: Let $a$ and $b$ be real constants such that the function $f$ defined by $f (x) = {x^{2} + 3 x + a b x + 2, x \leq 1, x > 1$ be differentiable on $R$ . Then, the value of $\int_{- 2}^{2} f (x) d x$ equals

Select Answer:

Visualized Solution

Understanding Differentiability

Condition for Continuity at $x = 1$

Substituting $x = 1$ for Continuity

First Equation: $a - b = - 2$

Differentiability Condition

Differentiating the Pieces

Solving for $b$

Solving for $a$

The Final Function

Setting up the Integral

Integrating the First Part

Evaluating First Integral Limits

Integrating the Second Part

Evaluating Second Integral Limits

Final Summation

The Sigma Insight: Relationship Between Continuity and Differentiability

The Art of the Smooth Transition

Phase 1

Phase 2

Phase 3

The Grand Finale

Similar Questions

Let the function, $f (x) = {- 3 a x^{2} - 2, a^{2} + b x, x < 1 x \geq 1$ be differentiable for all $x \in R$ where $a > 1,$ $b \in R$ . If the area of the region enclosed by $y = f (x)$ and the line $y = - 20$ is $α + β 3,$ $α, β \in Z$ then the value of $α + β$ is

If the function $f (x) = {\frac{1}{∣ x ∣} a x^{2} + 2 b, ∣ x ∣ \geq 2, ∣ x ∣ < 2$ is differentiable on $R$ , then $48 (a + b)$ is equal to_____

If $f (x) = {\frac{1}{∣ x ∣}, a x^{2} + b, ∣ x ∣ \geq 1 ∣ x ∣ < 1$ is differentiable at every point of the domain, then the values of $a$ and $b$ are respectively:

Let $α, β \in R$ be such that the function $f (x) = {2 α (x^{2} - 2) + 2 β x, (α + 3) x + (α - β), x < 1 x \geq 1$ be differentiable at all $x \in R$ . Then $34 (α + β)$ is equal to

If the function $g (x) = {k x + 1, m x + 2, 0 \leq x \leq 3 3 < x \leq 5$ is differentiable, then the value of $k + m$ is

If $∣ c ∣ \leq 1/2$ and $f (x)$ is a differentiable function at $x = 0$ given by $f (x) = ⎩ ⎨ ⎧ b sin^{- 1} (\frac{c + x}{2}), 1/2, \frac{e ^{a x /2} - 1}{x}, - \frac{1}{2} < x < 0 x = 0 0 < x < \frac{1}{2}$ Find the value of 'a'.

A function $f$ is defined on $[- 3, 3]$ as $f (x) = {min {∣ x ∣, 2 - x^{2}}, [∣ x ∣], - 2 \leq x \leq 2 2 < ∣ x ∣ \leq 3$ where $[x]$ denotes the greatest integer $\leq x$ . The number of points, where $f$ is not differentiable in $(- 3, 3)$ is

Let $f : [a, b] \to [1, \infty)$ be a continuous function and let $g : R \to R$ be defined as $g (x) = ⎩ ⎨ ⎧ 0, \int_{a}^{x} f (t) d t, \int_{a}^{b} f (t) d t, if x < a, if a \leq x \leq b; if x > b .$ then