Example Question - integration
Here are examples of questions we've helped users solve.
Definite Integral of the Square of Secant Function
<p>La función secante al cuadrado, \( \sec^2(x) \), es la derivada de la función tangente, \( \tan(x) \). Por lo tanto, para resolver esta integral definida, podemos aplicar la regla fundamental del cálculo que dice que la integral de la derivada de una función en un intervalo es igual a la diferencia de los valores de la función en los extremos del intervalo:</p> <p>\[ \int_{0}^{4} \sec^2 x \, dx = \tan(x) \Big|_{0}^{4} \]</p> <p>Calculamos los valores de \( \tan(x) \) en \( x = 4 \) y \( x = 0 \) y encontramos la diferencia:</p> <p>\[ = \tan(4) - \tan(0) \]</p> <p>Sin embargo, notarás que hay un problema al querer evaluar \( \tan(4) \), ya que 4 radianes está muy cerca de \( \frac{\pi}{2} \approx 1.5708 \) y la función tangente no está definida en \( \frac{\pi}{2} + n\pi \), donde \( n \) es cualquier número entero. Al sustituir \( x = 4 \) en la función tangente, obtendremos un valor no definido, lo que significa que la integral no converge en ese punto.</p> <p>Por ende, debido a la discontinuidad en \( x = 4 \), no podemos obtener un valor numérico real para esta integral definida con los límites dados de \( 0 \) a \( 4 \). Si se cometió un error al escribir los límites de integración y se trataba de un intervalo donde la función tangente es continua, los pasos a seguir para resolverla serían correctos.</p>
Solving an Improper Integral
<p>لحل التكامل الوارد في الصورة، أولاً، نبسط المقام كالتالي:</p> <p>\[ \int \frac{1}{x^{4} \sqrt{9 - x^{2}}} dx \]</p> <p>نستخدم التعويض التالي:</p> <p>\[ x = 3\sin(\theta) \]</p> <p>\[ dx = 3\cos(\theta) d\theta \]</p> <p>\[ 9 - x^{2} = 9 - 9\sin^{2}(\theta) = 9\cos^{2}(\theta) \]</p> <p>نعوض في التكامل الأصلي:</p> <p>\[ = \int \frac{3\cos(\theta)}{(3\sin(\theta))^{4} \sqrt{9\cos^{2}(\theta)}} d\theta \]</p> <p>\[ = \frac{1}{27} \int \frac{\cos(\theta)}{\sin^{4}(\theta)\cos(\theta)} d\theta \]</p> <p>\[ = \frac{1}{27} \int \csc^{4}(\theta) d\theta \]</p> <p>\[ = \frac{1}{27} \int \csc^{2}(\theta) \csc^{2}(\theta) d\theta \]</p> <p>نستخدم التكامل بالتجزئة أو جداول التكامل لإيجاد:</p> <p>\[ \int \csc^{2}(\theta) d\theta = - \cot(\theta) \]</p> <p>\[ \int \csc^{4}(\theta) d\theta = -\frac{1}{3} \cot(\theta)\csc^{2}(\theta) - \frac{2}{3} \int \csc^{2}(\theta) d\theta \]</p> <p>نحل التكامل ونعيد التعبير عن الناتج بدلالة \( x \) بعد العودة إلى التعويض الأول.</p> <p>هذا التكامل قد يتطلب تقنيات متقدمة في التعامل مع التكاملات الغير مناسبة والدوال المثلثية، ولكن من المهم العودة إلى التعويض الأصلي للوصول إلى النتيجة النهائية بدلالة \( x \).</p>
Solving a Differential Equation with Separation of Variables
\[ \begin{align*} &\text{Sea la ecuación diferencial dada por } \frac{1}{P(1-P)} dP = dt.\\ &\text{Separamos variables e integramos ambos lados:} \\ &\int \frac{1}{P(1-P)} dP = \int dt.\\ &\text{Para resolver la integral del lado izquierdo, realizamos una descomposición en fracciones parciales:} \\ &\frac{1}{P(1-P)} = \frac{A}{P} + \frac{B}{1-P}, \text{ donde se puede encontrar que } A = 1, B = 1.\\ &\text{Entonces:} \\ &\int \left( \frac{1}{P} + \frac{1}{1 - P} \right) dP = \int dt. \\ &\text{Integrando:} \\ &\ln|P| - \ln|1 - P| = t + C, \text{ donde } C \text{ es la constante de integración.} \\ &\text{Reescribimos la solución como:} \\ &\ln \left| \frac{P}{1 - P} \right| = t + C. \\ &\text{Exponenciando ambos lados para resolver para } P: \\ &e^{\ln \left| \frac{P}{1 - P} \right|} = e^{t + C}. \\ &\text{Simplificamos:} \\ &\left| \frac{P}{1 - P} \right| = e^C \cdot e^t, \\ &\text{donde } e^C \text{ puede considerarse como una nueva constante } K.\\ &\text{Por lo tanto, tenemos:} \\ &\frac{P}{1 - P} = \pm K e^t. \\ &\text{Si resolvemos para } P \text{ obtenemos:} \\ &P = \frac{\pm K e^t}{1 \pm K e^t}, \text{ que es la solución general de la ecuación diferencial.} \end{align*} \]
Solving a Differential Equation
<p>La ecuación diferencial proporcionada es separable y puede resolverse utilizando métodos de integración. La ecuación original es:</p> <p>\[\frac{1}{P(1-P)} \, dP = dt\]</p> <p>Separando las variables e integrando ambos lados, obtenemos:</p> <p>\[\int \frac{1}{P(1-P)} \, dP = \int dt\]</p> <p>Para resolver la integral del lado izquierdo, aplicamos fracciones parciales:</p> <p>\[\frac{1}{P(1-P)} = \frac{A}{P} + \frac{B}{1-P}\]</p> <p>Donde \( A(1-P) + BP = 1 \). Resolviendo para \( A \) y \( B \), obtenemos \( A = B = 1 \).</p> <p>Entonces:</p> <p>\[\int \left( \frac{1}{P} + \frac{1}{1-P} \right) \, dP = \int dt\]</p> <p>\[\ln|P| - \ln|1-P| = t + C\]</p> <p>Donde \( C \) es la constante de integración. Ahora aplicaremos la propiedad logarítmica de la diferencia de logaritmos para combinarlos en un solo logaritmo:</p> <p>\[\ln \left| \frac{P}{1-P} \right| = t + C\]</p> <p>Exponenciando ambos lados para despejar \( P \):</p> <p>\[\frac{P}{1-P} = e^{t+C}\]</p> <p>\[P = (1-P)e^{t+C}\]</p> <p>\[P = e^{t+C} - Pe^{t+C}\]</p> <p>\[P + Pe^{t+C} = e^{t+C}\]</p> <p>\[P(1 + e^{t+C}) = e^{t+C}\]</p> <p>\[P = \frac{e^{t+C}}{1 + e^{t+C}}\]</p> <p>Si definimos \( e^C \) como una nueva constante \( k \), obtenemos la solución general de la ecuación diferencial:</p> <p>\[P = \frac{ke^t}{1 + ke^t}\]</p>
Calculating the Average Value of a Quadratic Function on a Given Interval
لا يمكن تحديد حل السؤال بشكل دقيق لأن الصورة لا توضح الفترة الزمنية المطلوبة للتكامل. يرجى تقديم الفترة الزمنية المحددة لحساب قيمة المتوسط.
Calculating the Integral of a Power Function
<p>\[ \int f(x) \, dx = \int x^{-2} \, dx \]</p> <p>\[ = \int x^{-2} \, dx = \frac{x^{-2+1}}{-2+1} + C \]</p> <p>\[ = \frac{x^{-1}}{-1} + C \]</p> <p>\[ = -\frac{1}{x} + C \]</p>
Evaluation of an Improper Integral
<p>The image shows an improper integral of the form \(\int \sqrt{x} dx\). To solve this integral, we use the power rule for integration. The power rule states that \(\int x^n dx = \frac{x^{n+1}}{n+1} + C\), for any real number \(n\) not equal to -1, where \(C\) is the constant of integration.</p> <p>The integral can thus be solved as follows:</p> <p>\begin{align*} \int \sqrt{x} dx &= \int x^{1/2} dx \\ &= \frac{x^{(1/2) + 1}}{(1/2) + 1} + C \\ &= \frac{x^{3/2}}{3/2} + C \\ &= \frac{2}{3} x^{3/2} + C. \end{align*}</p> <p>Therefore, the solution to the integral is \(\frac{2}{3} x^{3/2} + C\).</p>
Finding the Integral of an Exponential Function
<p>\(\int 2 \cdot 5^x dx\)</p> <p>\(= 2 \int 5^x dx\)</p> <p>\(= 2 \cdot \frac{1}{\ln(5)} 5^x + C\)</p> <p>where \(C\) is the constant of integration.</p>
Integral Calculation for a Rational Function
<p>\( \int f(x) \,dx = \int \frac{3}{x} \,dx \)</p> <p>\( = 3 \int \frac{1}{x} \,dx \)</p> <p>\( = 3 \ln|x| + C \)</p> <p>where \( C \) is the constant of integration.</p>
Integration of a Power Function
<p>\(\int x^{-2} dx\)</p> <p>\(= \int x^{-2+1}(-2+1)^{-1} dx\)</p> <p>\(= \int x^{-1}(-1)^{-1} dx\)</p> <p>\(= -x^{-1+1}(-1+1)^{-1} + C\)</p> <p>\(= -1 \cdot x^0 + C\)</p> <p>\(= -1 + C\)</p> <p>\(= -1 + C\)</p>
Integration using Partial Fractions and Natural Logarithm Rules
这个问题涉及到不定积分。给定的积分是： ∫ x / (x^2 - 9) dx 我们可以使用分项积分法来解决这个积分。首先，通过将分子乘以一个常数然后加上或减去同样的常数，使得分子形式与分母的导数接近。因为x^2的导数是2x，我们可以将分子换成如下形式： x = (1/2) * (2x) 那么积分可以写作： ∫ [(1/2) * (2x)] / (x^2 - 9) dx 分子现在是分母导数的一半，可以使用自然对数的积分规则。我们可以在分子中再加上和减去9的一半（即4.5或9/2），以此构成分母减去常数的形式： ∫ [(1/2) * (2x) +/- 9/2] / (x^2 - 9) dx 这将积分分解为两部分： (1/2) * ∫ (2x / (x^2 - 9)) dx + (1/2) * ∫ (-9 / (x^2 - 9)) dx 第一部分是简单的自然对数积分，我们知道∫ du/u = ln|u| + C。第二部分，我们可以通过简单的代换来解决，u = x^2 - 9, du = 2xdx。 让我们解第一部分积分： (1/2) * ∫ (2x / (x^2 - 9)) dx = (1/2) * ln|x^2 - 9| + C 第二部分的积分，我们可以使用部分分式分解将其写成更简单的形式。我们知道x^2 - 9可以分解为(x-3)(x+3)，因此我们可以写： (1/2) * ∫ (-9 / (x^2 - 9)) dx = (1/2) * ∫ (-9 / ((x-3)(x+3))) dx 我们可以通过部分分式分解方法找到两个常数A和B，使得： -9 / ((x-3)(x+3)) = A/(x-3) + B/(x+3) 求出A和B以后，分别对两个简单的分式进行积分，结果会是两个对数函数。详细解的步骤如下： 1. 将-9 / ((x-3)(x+3))分解成A/(x-3) + B/(x+3)的形式。 2. 解出A和B的值。 3. 分别计算A/(x-3)和B/(x+3)的积分。 4. 将结果合并得到原积分的结果。 这个问题的关键在于处理积分中涉及的代数和对数函数的积分规则，以及如何将复杂表达式简化为可解形式。然而，请注意，给定的图像并未提供足够的信息来完成整个解题过程，因为部分分式分解所需的A和B值需要通过代数操作来解决。 总的来说，解决这个积分需要综合使用代数技巧、分项积分法以及对数积分规则。
Integration using Substitution
To solve the integral \[ \int \frac{dx}{1 + \sqrt{x}}, \] we can use a substitution. Let's set \( u = \sqrt{x} \), then \( u^2 = x \). Now, we differentiate both sides with respect to \( x \) to get the differential relationship between \( u \) and \( x \). This gives us \( 2u du = dx \), or \( dx = 2u du \). Substitute into the integral: \[ \int \frac{2u}{1 + u} du. \] Now, let's divide the numerator by the denominator to simplify: \[ \int \frac{2u}{1 + u} du = \int \left(2 - \frac{2}{1 + u}\right) du. \] Breaking this integral apart we get: \[ \int 2 du - \int \frac{2}{1 + u} du. \] Integrating term by term: \[ 2u - 2\ln|1 + u| + C. \] Now we substitute back to express in terms of \( x \): \[ 2\sqrt{x} - 2\ln|\sqrt{x} + 1| + C, \] where \( C \) is the constant of integration. Therefore, the solution to the integral is: \[ \int \frac{dx}{1 + \sqrt{x}} = 2\sqrt{x} - 2\ln|\sqrt{x} + 1| + C. \]
Solving Indefinite Integral of Polynomial Expression
The image shows the integral of a polynomial expression. I'll guide you through the steps to solve the indefinite integral: ∫(8x^3 - x^2 + 5x - 1)dx To integrate this polynomial term by term, you would use the power rule for integration. The power rule states that the integral of x^n dx is (1/(n+1)) * x^(n+1) + C, where C is the constant of integration. Applying this rule to each term, you would get: ∫8x^3 dx = 8 * ∫x^3 dx = 8 * (1/4) * x^4 = 2x^4 ∫(-x^2) dx = (-1) * ∫x^2 dx = (-1) * (1/3) * x^3 = -1/3x^3 ∫5x dx = 5 * ∫x dx = 5 * (1/2) * x^2 = 5/2x^2 ∫(-1) dx = -1 * ∫dx = -1 * x = -x Adding them all together, you get: 2x^4 - 1/3x^3 + 5/2x^2 - x + C Therefore, the integral of the given expression is: 2x^4 - 1/3x^3 + 5/2x^2 - x + C
Integrating Polynomials Using Power Rule
To solve the integral provided in the image: ∫(8x^3 - x^2 + 5x - 1)dx You need to integrate each term separately using the power rule for integration. The power rule states that the integral of x^n with respect to x is (x^(n+1))/(n+1), where n ≠ -1. Applying the power rule to each term: For 8x^3, ∫8x^3 dx = 8 * ∫x^3 dx = 8 * (x^(3+1))/(3+1) = 8 * (x^4)/4 = 2x^4 For -x^2, ∫(-x^2) dx = - ∫x^2 dx = - (x^(2+1))/(2+1) = - (x^3)/3 For 5x, ∫5x dx = 5 * ∫x dx = 5 * (x^(1+1))/(1+1) = 5 * (x^2)/2 = (5/2)x^2 For -1, ∫(-1) dx = - ∫1 dx = -x Now, combine all the integrated terms: ∫(8x^3 - x^2 + 5x - 1)dx = 2x^4 - (x^3)/3 + (5/2)x^2 - x + C Where C is the constant of integration.
Evaluating Definite Integrals of Polynomial Functions using Power Rule
This is a definite integral of a polynomial function, which we can evaluate using the Power Rule for integration. The Power Rule states that the integral of x^n is (x^(n+1)) / (n+1) plus a constant (C), for any real number n ≠ -1. So let's evaluate your integral: ∫(8x^3 - x^2 + 5x - 1) dx Integration term by term: ∫8x^3 dx = 8 * ∫x^3 dx = 8 * (x^(3+1) / (3+1)) = 2 * x^4 ∫-x^2 dx = - ∫x^2 dx = - (x^(2+1) / (2+1)) = - (1/3) * x^3 ∫5x dx = 5 * ∫x dx = 5 * (x^(1+1) / (1+1)) = (5/2) * x^2 ∫-1 dx = -x Combining these results: 2x^4 - (1/3)x^3 + (5/2)x^2 - x Without limits of integration given, we cannot evaluate for specific numbers. If there were limits, you would substitute the upper limit into the result for x, then subtract the result from substituting the lower limit. Since no limits are provided, this is the indefinite integral result, and don't forget to add the constant of integration C: 2x^4 - (1/3)x^3 + (5/2)x^2 - x + C
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