Resourcesdown chevron
Testimonials
Plans
Sign in
Sign up
Resourcesright arrow
Testimonials
Plans
Bytelearn - cat image with glassesAI tutor

Welcome to Bytelearn!

Let’s check out your problem:

Integrate. int_(0)^(1)((3x^(3)-x^(2)+2x-4))/(sqrt(x^(2)-3x+2))dx

Integrate.\newline∫01(3x3−x2+2x−4)x2−3x+2dx \int_{0}^{1} \frac{\left(3 x^{3}-x^{2}+2 x-4\right)}{\sqrt{x^{2}-3 x+2}} d x

View step-by-step helpView step-by-step help
Homeright chevron icon
Math Problemsright chevron icon
Calculusright chevron icon
Evaluate definite integrals using the chain ruleright chevron icon

Full solution

Q. Integrate.\newline∫01(3x3−x2+2x−4)x2−3x+2dx \int_{0}^{1} \frac{\left(3 x^{3}-x^{2}+2 x-4\right)}{\sqrt{x^{2}-3 x+2}} d x
  1. Simplify Denominator: Simplify the denominator of the integrand.\newlineThe denominator is x2−3x+2\sqrt{x^2 - 3x + 2}. We can factor the quadratic inside the square root to see if it simplifies further.\newlinex2−3x+2=(x−1)(x−2)x^2 - 3x + 2 = (x - 1)(x - 2)\newlineSo, x2−3x+2=(x−1)(x−2)\sqrt{x^2 - 3x + 2} = \sqrt{(x - 1)(x - 2)}
  2. Check Domain: Check the domain of the integrand.\newlineSince we have a square root in the denominator, we need to ensure that the expression inside the square root is non-negative for all xx in the interval [0,1][0, 1].\newline(x−1)(x−2)≤0(x - 1)(x - 2) \leq 0 for xx in [0,1][0, 1] because both (x−1)(x - 1) and (x−2)(x - 2) are negative in this interval. Therefore, the square root is well-defined and real.
  3. Set Up Integral: Set up the integral.\newlineWe are looking to evaluate the integral from 00 to 11 of the function (3x3−x2+2x−4)/x2−3x+2(3x^3 - x^2 + 2x - 4) / \sqrt{x^2 - 3x + 2}.\newline∫013x3−x2+2x−4(x−1)(x−2)dx\int_{0}^{1} \frac{3x^3 - x^2 + 2x - 4}{\sqrt{(x - 1)(x - 2)}} dx
  4. Simplify Integrand: Attempt to simplify the integrand.\newlineThe integrand does not simplify easily, and there is no obvious substitution that will cancel the terms in the numerator with the denominator. This suggests that we may need to use a more advanced technique, such as partial fractions or trigonometric substitution, to evaluate the integral.
  5. Trig Substitution: Recognize the need for a trigonometric substitution. Given the form of the denominator, (x−1)(x−2)\sqrt{(x - 1)(x - 2)}, a trigonometric substitution might be appropriate. However, the form does not match the standard forms for trigonometric substitution directly. We need to complete the square in the denominator to see if a substitution becomes apparent.
  6. Complete Square: Complete the square in the denominator.\newlineTo complete the square for x2−3x+2x^2 - 3x + 2, we take the coefficient of xx, which is −3-3, divide it by 22 to get −32-\frac{3}{2}, and square it to get 94\frac{9}{4}. We then add and subtract 94\frac{9}{4} inside the square root to complete the square.\newlinex2−3x+2=x2−3x+94−94+2=(x−32)2−14\sqrt{x^2 - 3x + 2} = \sqrt{x^2 - 3x + \frac{9}{4} - \frac{9}{4} + 2} = \sqrt{(x - \frac{3}{2})^2 - \frac{1}{4}}
  7. Apply Substitution: Apply trigonometric substitution.\newlineNow that we have the form (x−32)2−14\sqrt{(x - \frac{3}{2})^2 - \frac{1}{4}}, we can use the substitution x−32=12sec⁡(θ)x - \frac{3}{2} = \frac{1}{2}\sec(\theta) to simplify the square root.\newlineLet x−32=12sec⁡(θ)x - \frac{3}{2} = \frac{1}{2}\sec(\theta), then dx=12sec⁡(θ)tan⁡(θ)dθdx = \frac{1}{2}\sec(\theta)\tan(\theta)d\theta.
  8. Change Limits: Change the limits of integration.\newlineWhen we change the variable of integration from xx to θ\theta, we also need to change the limits of integration. For x=0x = 0, we have 0−32=(12)sec⁡(θ)0 - \frac{3}{2} = (\frac{1}{2})\sec(\theta), which gives sec⁡(θ)=−3\sec(\theta) = -3. This is not possible since sec⁡(θ)\sec(\theta) is always positive or negative infinity, indicating a potential error in our substitution choice. We need to reassess our substitution strategy.

More problems from Evaluate definite integrals using the chain rule

Question
Consider the following problem:\newlineThe water level under a bridge is changing at a rate of r(t)=40sin⁡(πt6) r(t)=40 \sin \left(\frac{\pi t}{6}\right) centimeters per hour (where t t is the time in hours). At time t=3 t=3 , the water level is 9191 centimeters. By how much does the water level change during the 4th  4^{\text {th }} hour?\newlineWhich expression can we use to solve the problem?\newlineChoose 11 answer:\newline(A) ∫34r(t)dt \int_{3}^{4} r(t) d t \newline(B) ∫04r(t)dt \int_{0}^{4} r(t) d t \newline(C) ∫45r(t)dt \int_{4}^{5} r(t) d t \newline(D) ∫44r(t)dt \int_{4}^{4} r(t) d t
Get tutor helpright-arrow

Posted 8 months ago

Question
The function s(t) s(t) gives the number of students enrolled in a school by time t t (in years).\newlineWhat does ∫1518s′(t)dt=20 \int_{15}^{18} s^{\prime}(t) d t=20 mean?\newlineChoose 11 answer:\newline(A) The rate of change of enrollment is 2020 students per year more in year 1818 than it was in year 1515.\newline(B) There were 2020 more students enrolled in year 1818 than in year 1515 .\newline(C) Between years 1515 and 1818, the cumulative number of years of schooling of all of the enrolled students is 2020 years.\newline(D) There are 2020 students enrolled in year 1818.
Get tutor helpright-arrow

Posted 7 months ago

Question
A water tank is filled at a rate of r(t) r(t) liters per minute (where t t is the time in minutes).\newlineWhat does ∫17r′(t)dt \int_{1}^{7} r^{\prime}(t) d t represent?\newlineChoose 11 answer:\newline(A) The rate at which the tank was filled at t=7 t=7 .\newlineB) The average rate of filling between t=1 t=1 and t=7 t=7 .\newline(C) The change in the rate of filling between t=1 t=1 and t=7 t=7 .\newline(D) The amount of water filled between t=1 t=1 and t=7 t=7 .
Get tutor helpright-arrow

Posted 8 months ago

Question
The base of a solid is the region enclosed by the graphs of y=sin⁡(x) y=\sin (x) and y=4−x y=4-\sqrt{x} , between x=2 x=2 and x=7 x=7 .\newlineCross sections of the solid perpendicular to the x x -axis are rectangles whose height is 2x 2 x .\newlineWhich one of the definite integrals gives the volume of the solid?\newlineChoose 11 answer:\newline(A) ∫27[(4−x)2−sin⁡2(x)]dx \int_{2}^{7}\left[(4-\sqrt{x})^{2}-\sin ^{2}(x)\right] d x \newline(B) ∫27[sin⁡(x)+x−4]⋅2xdx \int_{2}^{7}[\sin (x)+\sqrt{x}-4] \cdot 2 x d x \newline(C) ∫27[sin⁡2(x)−(4−x)2]dx \int_{2}^{7}\left[\sin ^{2}(x)-(4-\sqrt{x})^{2}\right] d x \newline(D) ∫27[4−x−sin⁡(x)]⋅2xdx \int_{2}^{7}[4-\sqrt{x}-\sin (x)] \cdot 2 x d x
Get tutor helpright-arrow

Posted 8 months ago

Question
Evaluate the integral.\newline∫−2x3cos⁡(3x)dx \int-2 x^{3} \cos (3 x) d x \newlineAnswer:
Get tutor helpright-arrow

Posted 9 months ago

Question
Evaluate the integral.\newline∫−4x233xdx \int-4 x^{2} 3^{3 x} d x \newlineAnswer:
Get tutor helpright-arrow

Posted 9 months ago

Question
Evaluate the integral.\newline∫−4x2e4xdx \int-4 x^{2} e^{4 x} d x \newlineAnswer:
Get tutor helpright-arrow

Posted 9 months ago

Question
∑n=1∞3nn2n!\sum_{n=1}^{\infty}\frac{3^{n}n^{2}}{n!}
Get tutor helpright-arrow

Posted 8 months ago

Question
f)lim⁡n→+∞[2n+(35)n] f) \lim_{n \to +\infty}\left[\frac{2}{n}+\left(\frac{3}{5}\right)^n\right]
Get tutor helpright-arrow

Posted 8 months ago

Question
∫1cos⁡(x)2 dx\int \frac{1}{\cos(x)^{2}}\,dx
Get tutor helpright-arrow

Posted 8 months ago

View all (198)

Related topics

Algebra - Order of OperationsAlgebra - Distributive Property`X` and `Y` AxesGeometry - Scalene TriangleCommon MultipleGeometry - Quadrant