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Math  /  Calculus

Question∫x4−2x3+x2dx\int \sqrt{x^{4}-2 x^{3}+x^{2}} d x

Studdy Solution

STEP 1

1. The integrand x4−2x3+x2\sqrt{x^{4} - 2x^{3} + x^{2}} is a rational function under the square root.
2. The integrand can potentially be simplified or transformed to facilitate the integration process.
3. Standard integration techniques, such as substitution or trigonometric identities, may be used.

STEP 2

1. Simplify the expression inside the square root.
2. Determine an appropriate substitution to simplify the integral.
3. Perform the substitution and simplify the resulting integral.
4. Integrate the simplified expression.
5. Substitute back to the original variable.
6. Simplify the final expression to obtain the result.

STEP 3

Simplify the expression inside the square root.
x4−2x3+x2=x2(x2−2x+1) \sqrt{x^4 - 2x^3 + x^2} = \sqrt{x^2 (x^2 - 2x + 1)}

STEP 4

Factor the quadratic expression inside the square root.
x2−2x+1=(x−1)2 x^2 - 2x + 1 = (x - 1)^2

STEP 5

Substitute back into the square root expression.
x2(x2−2x+1)=x2(x−1)2=x2⋅(x−1)2=∣x∣⋅∣x−1∣ \sqrt{x^2 (x^2 - 2x + 1)} = \sqrt{x^2 (x - 1)^2} = \sqrt{x^2} \cdot \sqrt{(x - 1)^2} = |x| \cdot |x - 1|

STEP 6

Break down the absolute value expression into cases based on the values of xx.
For x≥1x \geq 1: ∣x∣=xand∣x−1∣=x−1 |x| = x \quad \text{and} \quad |x-1| = x-1 For 0≤x<10 \leq x < 1: ∣x∣=xand∣x−1∣=1−x |x| = x \quad \text{and} \quad |x-1| = 1-x

STEP 7

Consider the integral over the interval x≥1x \geq 1 first.
∫1∞x(x−1) dx=∫1∞(x2−x) dx \int_{1}^{\infty} x(x - 1) \, dx = \int_{1}^{\infty} (x^2 - x) \, dx

STEP 8

Integrate the expression x2−xx^2 - x for x≥1x \geq 1.
∫(x2−x) dx=∫x2 dx−∫x dx \int (x^2 - x) \, dx = \int x^2 \, dx - \int x \, dx

STEP 9

Evaluate the integrals separately.
∫x2 dx=x33+C \int x^2 \, dx = \frac{x^3}{3} + C ∫x dx=x22+C \int x \, dx = \frac{x^2}{2} + C

STEP 10

Combine the results of the integrals.
∫(x2−x) dx=x33−x22+C \int (x^2 - x) \, dx = \frac{x^3}{3} - \frac{x^2}{2} + C

STEP 11

Simplify and evaluate at the bounds for x≥1x \geq 1.
[x33−x22]1∞ \left[ \frac{x^3}{3} - \frac{x^2}{2} \right]_{1}^{\infty}

STEP 12

Evaluate the simplified integral at the bounds.
lim⁡b→∞(b33−b22)−(133−122) \lim_{b \to \infty} \left( \frac{b^3}{3} - \frac{b^2}{2} \right) - \left( \frac{1^3}{3} - \frac{1^2}{2} \right)

STEP 13

Noticing that the integral diverges as b→∞b \to \infty since b33\frac{b^3}{3} grows without bound.
Therefore, the integral does not converge over x≥1x \geq 1.
Consider the integral over the interval 0≤x<10 \leq x < 1.
∫01x(1−x) dx=∫01(x−x2) dx \int_{0}^{1} x(1 - x) \, dx = \int_{0}^{1} (x - x^2) \, dx

STEP 14

Integrate the expression x−x2x - x^2 for 0≤x<10 \leq x < 1.
∫(x−x2) dx=∫x dx−∫x2 dx \int (x - x^2) \, dx = \int x \, dx - \int x^2 \, dx

STEP 15

Evaluate the integrals separately.
∫x dx=x22+C \int x \, dx = \frac{x^2}{2} + C ∫x2 dx=x33+C \int x^2 \, dx = \frac{x^3}{3} + C

STEP 16

Combine the results of the integrals.
∫(x−x2) dx=x22−x33+C \int (x - x^2) \, dx = \frac{x^2}{2} - \frac{x^3}{3} + C

STEP 17

Simplify and evaluate at the bounds for 0≤x<10 \leq x < 1.
[x22−x33]01 \left[ \frac{x^2}{2} - \frac{x^3}{3} \right]_{0}^{1}

STEP 18

Evaluate the simplified integral at the bounds.
(122−133)−(022−033)=12−13=36−26=16 \left( \frac{1^2}{2} - \frac{1^3}{3} \right) - \left( \frac{0^2}{2} - \frac{0^3}{3} \right) = \frac{1}{2} - \frac{1}{3} = \frac{3}{6} - \frac{2}{6} = \frac{1}{6}
The final solution for the integral over 0≤x<10 \leq x < 1 is:
∫01x4−2x3+x2 dx=16 \int_{0}^{1} \sqrt{x^4 - 2x^3 + x^2} \, dx = \frac{1}{6}

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