Example Question - solving integral
Here are examples of questions we've helped users solve.
Solving Absolute Value Integral
To solve the integral \(\int_{-1}^{2} |x - x^2|\, dx\), we need to consider when the expression inside the absolute value, \(x - x^2\), is positive or negative. First, let's find the points where \(x - x^2 = 0\): \(x - x^2 = x(1 - x) = 0\) This gives us two points \(x = 0\) and \(x = 1\). Now we divide the interval \([-1, 2]\) based on these points and analyze the sign of the expression \(x - x^2\) in each subinterval: 1. For \(x \in [-1, 0]\), the expression is always negative since both \(x\) and \(x^2\) are non-positive and \(x^2\) is greater in magnitude than \(x\). 2. For \(x \in [0, 1]\), the expression is positive since \(x\) is greater than or equal to \(x^2\). 3. For \(x \in [1, 2]\), the expression is negative since \(x^2\) is greater than \(x\). Thus, we need to change the sign of the expression in the intervals where it is negative, which gives us: \[ \int_{-1}^{0} -(x - x^2)\, dx + \int_{0}^{1} (x - x^2)\, dx + \int_{1}^{2} -(x - x^2)\, dx \] Now we can integrate each term separately: 1. \(\int_{-1}^{0} -(x - x^2)\, dx = -\frac{1}{2} x^2 + \frac{1}{3}x^3\Big|_{-1}^{0}\) 2. \(\int_{0}^{1} (x - x^2)\, dx = \frac{1}{2} x^2 - \frac{1}{3}x^3\Big|_{0}^{1}\) 3. \(\int_{1}^{2} -(x - x^2)\, dx = -\frac{1}{2} x^2 + \frac{1}{3}x^3\Big|_{1}^{2}\) Now, evaluate each integral: 1. \(-\frac{1}{2} (0)^2 + \frac{1}{3}(0)^3 - (-\frac{1}{2} (-1)^2 + \frac{1}{3}(-1)^3) = 0 - (-\frac{1}{2} + \frac{1}{3}) = \frac{1}{3} - \frac{1}{2} = -\frac{1}{6}\) 2. \(\frac{1}{2} (1)^2 - \frac{1}{3}(1)^3 - (\frac{1}{2} (0)^2 - \frac{1}{3}(0)^3) = \frac{1}{2} - \frac{1}{3} - 0 = \frac{1}{6}\) 3. \(-\frac{1}{2} (2)^2 + \frac{1}{3}(2)^3 - (-\frac{1}{2} (1)^2 + \frac{1}{3}(1)^3) = -2 + \frac{8}{3} - (-\frac{1}{2} + \frac{1}{3}) = -2 + \frac{8}{3} + \frac{1}{6} = -2 + \frac{8}{3} + \frac{1}{6}\) To add \(-2 + \frac{8}{3} + \frac{1}{6}\), first find a common denominator, which is 6: \(-2 + \frac{8 \cdot 2}{3 \cdot 2} + \frac{1}{6} = -2 + \frac{16}{6} + \frac{1}{6} = -2 + \frac{17}{6}\) \(-2\) is equivalent to \(- \frac{12}{6}\), so we have: \(-\frac{12}{6} + \frac{17}{6} = \frac{5}{6}\) Adding the results of all the terms: \(-\frac{1}{6} + \frac{1}{6} + \frac{5}{6} = \frac{5}{6}\) So, the value of the integral \(\int_{-1}^{2} |x - x^2|\, dx\) is \(\frac{5}{6}\).
Solving an Integral Step by Step
The image shows an integral that needs to be solved. Let's solve the integral step by step: The integral is \( \int (8x^3 - x^2 + 5x - 1) \,dx \). To solve it, we will integrate each term separately: 1. The integral of \( 8x^3 \) with respect to x is \( \frac{8}{4}x^4 \) or \( 2x^4 \). 2. The integral of \( -x^2 \) with respect to x is \( -\frac{1}{3}x^3 \). 3. The integral of \( 5x \) with respect to x is \( \frac{5}{2}x^2 \). 4. The integral of \( -1 \) with respect to x is \( -x \). Putting it all together, we get: \[ \int (8x^3 - x^2 + 5x - 1) \,dx = 2x^4 - \frac{1}{3}x^3 + \frac{5}{2}x^2 - x + C \] where \( C \) is the constant of integration.
Solving an Integral using Power Rule for Integration
The image depicts an integral that needs to be solved. The integral is: ∫(8x^3 - x^2 + 5x - 1) dx To solve this integral, you can integrate each term separately. The integral of a sum/difference is the sum/difference of the integrals, and the power rule for integration states that ∫x^n dx = x^(n+1)/(n+1) + C, where C is the constant of integration. Applying this rule to each term gives us: ∫8x^3 dx - ∫x^2 dx + ∫5x dx - ∫1 dx Now integrate each term: 8 * ∫x^3 dx - ∫x^2 dx + 5 * ∫x dx - ∫1 dx = 8 * (x^4/4) - (x^3/3) + 5 * (x^2/2) - x + C = 2x^4 - (1/3)x^3 + (5/2)x^2 - x + C This is the antiderivative of the given function. So the solution to the integral is: 2x^4 - (1/3)x^3 + (5/2)x^2 - x + C
Solving Integral of Polynomial Function
The image shows an integral that needs to be solved. The expression within the integral is a polynomial function: ∫(8x^3 - x^2 + 5x - 1) dx To solve this, we integrate each term of the polynomial separately with respect to x. The integral of a constant times a power of x is the constant times the fraction of the incremented power of x divided by the new exponent. The integral of a constant is just the constant times x. Here's how this integral is computed: ∫8x^3 dx = 2x^4 (because you add 1 to the exponent and then divide by the new exponent) ∫-x^2 dx = -x^3/3 (same principle) ∫5x dx = 5x^2/2 (again, same principle) ∫-1 dx = -x Putting it all together, the antiderivative of the given function is: 2x^4 - x^3/3 + 5x^2/2 - x + C where C is the constant of integration.
CamTutor
In regards to math, we are professionals.
Get In Touch
Email: camtutor.ai@gmail.com