If f(x) = ∫_(x²)^(x²+1) e^(-t²) dt, then f(x) increases in (a) (-2, 2) (b) no value of x (c) (0, ∞)

Q: Wait, why can't we just integrate $e^{-t^2}$ and then differentiate?

That is a very common instinct! However, the function $e^{-t^2}$ does not have an elementary antiderivative. We cannot express the integral in terms of standard functions like polynomials or logarithms. This is exactly why the Leibniz Rule is so powerful—it allows us to find the derivative of the integral without ever needing to perform the integration itself.

Q: Why is $e^{-(x^2+1)^2} < e^{-x^4}$? I thought exponents were always positive.

Remember that the function $y = e^{-u}$ is a strictly decreasing function. This means that if you have two inputs $u_1$ and $u_2$ where $u_1 > u_2$, then $e^{-u_1} < e^{-u_2}$. In our case, $(x^2+1)^2$ is clearly larger than $x^4$ for all real $x$. Therefore, $e$ raised to the negative of the larger value must be smaller than $e$ raised to the negative of the smaller value.

Q: Why does $f'(x) > 0$ lead to $x < 0$?

Great catch! We found that $f'(x) = 2x \cdot (\text{Negative Quantity})$. For the product of two numbers to be positive, they must have the same sign. Since the bracketed term is always negative, the only way for the entire expression to be positive is if $2x$ is also negative. Thus, $x < 0$ is the required condition.

Question

If $f(x) = \int_{x^2}^{x^2+1} e^{-t^2} dt$, then $f(x)$ increases in

Options:

(A) (-2, 2)

(B) no value of x

(C) $(0, \infty)$

(D) $(-\infty, 0)$

Sigma Percentile · Accepted Answer

### Analyzing the Setup

Imagine you are standing before the graph of the Gaussian function, $y = e^{-t^2}$. It is a beautiful, symmetric bell curve, centered at the origin.

Now, consider the function $f(x) = \int_{x^2}^{x^2+1} e^{-t^2} dt$. Geometrically, this function represents the area under that bell curve, trapped between two vertical lines at $t = x^2$ and $t = x^2 + 1$.

As you vary $x$, this 'window' of integration slides along the $t$-axis. Because $x^2$ is always non-negative, our window is forever confined to the right side of the $y$-axis. To determine if this area grows or shrinks as $x$ increases, we must calculate the derivative $f'(x)$.

### The Magic of Leibniz

When we face an integral where the limits are functions of $x$, the Leibniz Rule provides a systematic way to differentiate under the integral sign:

$$f'(x) = \frac{d}{dx} \int_{u(x)}^{v(x)} g(t) dt = g(v(x)) \cdot v'(x) - g(u(x)) \cdot u'(x)$$

Here, our integrand is $g(t) = e^{-t^2}$, our upper limit is $v(x) = x^2 + 1$, and our lower limit is $u(x) = x^2$. By applying this rule, we transform a daunting integral into a manageable algebraic expression.

### The Algebraic Grind

For the upper limit, we have $e^{-(x^2+1)^2}$ multiplied by the derivative of $x^2+1$, which is $2x$. For the lower limit, we have $e^{-(x^2)^2}$ multiplied by the derivative of $x^2$, which is $2x$.

Putting it all together, we get:

$$f'(x) = 2x \cdot e^{-(x^2+1)^2} - 2x \cdot e^{-x^4}$$

Factoring out the $2x$, we arrive at:

$$f'(x) = 2x \left[ e^{-(x^2+1)^2} - e^{-x^4} ight]$$

This expression is the heartbeat of our problem. To find where the function increases, we need $f'(x) > 0$.

### The Conceptual Pivot

We must now determine the sign of the bracketed term by comparing $e^{-(x^2+1)^2}$ and $e^{-x^4}$. For any real $x$, it is clear that $x^2+1 > x^2$, which implies $(x^2+1)^2 > x^4$.

Since the function $y = e^{-u}$ is strictly decreasing, a larger input $u$ results in a smaller output $e^{-u}$. Because $(x^2+1)^2$ is the larger input, $e^{-(x^2+1)^2}$ must be smaller than $e^{-x^4}$.

Consequently, the difference $e^{-(x^2+1)^2} - e^{-x^4}$ is **strictly negative** for all $x$.

### The Final Revelation

We are left with the inequality:

$$2x \cdot (	ext{Negative Quantity}) > 0$$

For this product to be positive, $2x$ must be negative. Dividing by $2$, we find $x < 0$.

Thus, the function $f(x)$ increases precisely when $x$ is in the interval **$(-\infty, 0)$**. You have navigated the geometry, mastered the Leibniz Rule, and conquered the sign analysis to reach this conclusion.

Animated Solution for Mathematics - Differentiation: If $f (x) = \int_{x^{2}}^{x^{2} + 1} e^{- t^{2}} d t$ , then $f (x)$ increases in

Select Answer:

Visualized Solution

Visualizing the Integral $f (x)$

Condition for Monotonicity: $f^{'} (x) > 0$

The Leibniz Rule

Identifying Components for Leibniz Rule

Differentiating the Upper Limit

Differentiating the Lower Limit

Forming the Expression for $f^{'} (x)$

Analyzing the Exponents

Applying the Decreasing Function $e^{- t}$

Sign of the Bracketed Term

Solving the Inequality $f^{'} (x) > 0$

Final Interval of Increase

The Sigma Insight: Monotonicity

Analyzing the Setup

The Magic of Leibniz

The Algebraic Grind

The Conceptual Pivot

The Final Revelation

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Animated Solution for Mathematics - Differentiation: .math-text-wrapper .katex { font-size: 1.05em !important; } .math-text-wrapper .katex-display { margin: 1rem 0 !important; } If f(x)=∫x2x2+1​e−t2dt, then f(x) increases in