Question

Find $\frac{d}{dx}{\int }_{3-x}^{x^2}(x-t)dt$ both by evaluating the integral first and by differentiating first.

Short answer

The derivative is $(3x^2-2x^3-6+3x)$ or $2x^2-2x^3+2x-3$.

Step-by-step solution

- Evaluate the integral

Consider the integral $\frac{d}{dx}{\int }_{3-x}^{x^2}(x-t)dt$. First, evaluate the inner integral with respect to $t$.$$I(x)={\int }_{3-x}^{x^2}(x-t)dt$$This results in:$$I(x)={[xt-\frac{t^2}{2}]}_{3-x}^{x^2}$$

- Apply the limits

Substitute the upper and lower limits of the integral.For the upper limit $t=x^2$:$$I_{upper}=x(x^2)-\frac{(x^2{)}^2}{2}=x^3-\frac{x^4}{2}$$For the lower limit $t=3-x$:$$I_{lower}=x(3-x)-\frac{(3-x{)}^2}{2}=3x-x^2-\frac{9-6x+x^2}{2}=3x-x^2-\frac{9}{2}+3x-\frac{x^2}{2}$$

- Simplify the evaluated integral

Combine the results from the upper and lower limits to simplify:$$I(x)=x^3-\frac{x^4}{2}-(6x-\frac{3x^2}{2}-\frac{9}{2})$$Finally, we get:$$I(x)=x^3-\frac{x^4}{2}-6x+\frac{3x^2}{2}+\frac{9}{2}$$

- Differentiate the result

Differentiate $I(x)$ with respect to $x$ to find $\frac{d}{dx}I(x)$:$$\frac{d}{dx}\text{(}{x}^3-\frac{x^4}{2}-6x+\frac{3x^2}{2}+\frac{9}{2}\text{)}=3x^2-2x^3-6+3x$$

- Differentiate first using Leibniz rule

Use the Leibniz rule to differentiate under the integral sign:$$\frac{d}{dx}{\int }_{a(x)}^{b(x)}f(x,t)dt=f(x,b(x))\frac{db(x)}{dx}-f(x,a(x))\frac{da(x)}{dx}+{\int }_{a(x)}^{b(x)}\frac{\partial }{\partial x}f(x,t)dt$$For our integral, $\frac{d}{dx}{\int }_{3-x}^{x^2}(x-t)dt$:$$\frac{d}{dx}{\int }_{3-x}^{x^2}(x-t)dt=(x-x^2)\frac{d(x^2)}{dx}-(x-(3-x))\frac{d(3-x)}{dx}$$

- Simplify using derivatives of limits

Differentiate the limits and substitute:$$(x-x^2)\times 2x-(x-3+x)\times (-1)=(x-x^2)\times 2x-(2x-3)\times (-1)$$Simplify to get:$$(2x^2-2x^3)+2x-3$$

Key concepts

Differentiation

Differentiation is a fundamental idea in calculus that deals with finding the rate at which a function changes at any given point. In simple terms, differentiation helps us determine the slope or steepness of a curve. This is achieved through the derivative, which measures how a function's output changes with respect to changes in its input. For the function $f(x)$, its derivative is denoted as $f^{\prime }(x)$ or $\frac{d}{dx}f(x)$.
When working with differentiation, you might come across various rules like the product rule, quotient rule, and chain rule, which help simplify the process of finding derivatives for complex functions. These rules make it easier to tackle real-life problems involving motion, growth rates, and optimization.
In the context of our exercise, we used differentiation to find the final rate of change of the given integral after evaluating it first. This step involves applying the basic rules of differentiation to the resulting polynomial expression.

Integral Calculus

Integral calculus is another core area of calculus that focuses on aggregation, accumulation, and the area under curves. Integrals can be viewed as the inverse operation of differentiation. If differentiation breaks functions apart, integrating them puts them back together. There are two main types of integrals: definite and indefinite.
A definite integral, like the one in our exercise, computes the accumulation of quantities over a specific interval. It's denoted as ${\int }_a^{b}f(t)dt$, where $a$ and $b$ are the lower and upper bounds, respectively. This integral measures the net area between the function $f(t)$ and the x-axis, from $t=a$ to $t=b$.
In the exercise, we evaluated the definite integral ${\int }_{3-x}^{x^2}(x-t)dt$. We processed it by handling the inner integrand and then applying the limits to find the accumulated change over the interval.

Fundamental Theorem of Calculus

The Fundamental Theorem of Calculus bridges the gap between differentiation and integration. This theorem consists of two main parts.
The first part states: If $F(x)$ is an antiderivative of $f(x)$, then the definite integral of $f(t)$ from $a$ to $b$ is given by $F(b)-F(a)$. Mathematically, $${\int }_a^{b}f(t)dt=F(b)-F(a)$$
The second part states: If $F(x)$ is defined by $F(x)={\int }_a^{x}f(t)dt$, then $F^{\prime }(x)=f(x)$. This means differentiation and integration are inverse processes.
For our exercise, this theorem allows us to move between evaluating an integral directly and using derivatives to handle integrals with changing limits. The Leibniz rule, which is part of what we used in our problem, applies this second part by integrating first and then differentiating the resulting expression.

Derivatives of Integrals

Derivatives of integrals come into play when we deal with integrals that have variable limits. This scenario is exactly where the Leibniz rule becomes crucial. The Leibniz rule helps us find the derivative of an integral where both the integrand and the limits of integration change with respect to a variable.
The Leibniz rule is expressed as:
\ [ \frac{d}{dx} \int_{a(x)}^{b(x)} f(x, t) \, dt = f(x, b(x)) \frac{d b(x)}{d x} - f(x, a(x)) \frac{d a(x)}{d x} + \int_{a(x)}^{b(x)} \frac{\partial}{\partial x} f(x, t) \, dt \ ]
In our problem, we applied this rule to transform the given integral into functions of $x$ and then found their derivatives. The terms $(x-x^2)\cdot 2x$ and $(x-3+x)\cdot (-1)$ in our computation came directly from differentiating the integrand and the variable limits according to the Leibniz rule.
Understanding derivatives of integrals using the Leibniz rule can significantly simplify complex calculus problems where limits are not constant.