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Math  /  Calculus

Questionf) ∫13x2ln⁡xdx\int_{1}^{3} x^{2} \ln x d x

Studdy Solution

STEP 1

1. The integral ∫13x2ln⁡x dx\int_{1}^{3} x^{2} \ln x \, dx requires integration by parts.
2. We will use the integration by parts formula: ∫u dv=uv−∫v du\int u \, dv = uv - \int v \, du.
3. We need to choose uu and dvdv appropriately.

STEP 2

1. Choose uu and dvdv.
2. Differentiate uu to find dudu and integrate dvdv to find vv.
3. Apply the integration by parts formula.
4. Evaluate the definite integral.

STEP 3

Choose uu and dvdv. For this problem, let:
u=ln⁡x u = \ln x dv=x2 dx dv = x^2 \, dx

STEP 4

Differentiate uu to find dudu and integrate dvdv to find vv:
du=1x dx du = \frac{1}{x} \, dx v=∫x2 dx=x33 v = \int x^2 \, dx = \frac{x^3}{3}

STEP 5

Apply the integration by parts formula ∫u dv=uv−∫v du\int u \, dv = uv - \int v \, du:
∫x2ln⁡x dx=(ln⁡x⋅x33)−∫(x33⋅1x) dx \int x^2 \ln x \, dx = \left( \ln x \cdot \frac{x^3}{3} \right) - \int \left( \frac{x^3}{3} \cdot \frac{1}{x} \right) \, dx
Simplify the integral:
=x3ln⁡x3−∫x23 dx = \frac{x^3 \ln x}{3} - \int \frac{x^2}{3} \, dx

STEP 6

Evaluate the remaining integral:
∫x23 dx=13∫x2 dx=13⋅x33=x39 \int \frac{x^2}{3} \, dx = \frac{1}{3} \int x^2 \, dx = \frac{1}{3} \cdot \frac{x^3}{3} = \frac{x^3}{9}
Substitute back into the expression:
∫x2ln⁡x dx=x3ln⁡x3−x39 \int x^2 \ln x \, dx = \frac{x^3 \ln x}{3} - \frac{x^3}{9}

STEP 7

Evaluate the definite integral from 1 to 3:
[x3ln⁡x3−x39]13 \left[ \frac{x^3 \ln x}{3} - \frac{x^3}{9} \right]_{1}^{3}
Calculate at the upper limit x=3x = 3:
33ln⁡33−339=27ln⁡33−279=9ln⁡3−3 \frac{3^3 \ln 3}{3} - \frac{3^3}{9} = \frac{27 \ln 3}{3} - \frac{27}{9} = 9 \ln 3 - 3
Calculate at the lower limit x=1x = 1:
13ln⁡13−139=0−19=−19 \frac{1^3 \ln 1}{3} - \frac{1^3}{9} = 0 - \frac{1}{9} = -\frac{1}{9}
Subtract the lower limit evaluation from the upper limit evaluation:
(9ln⁡3−3)−(−19)=9ln⁡3−3+19 (9 \ln 3 - 3) - \left(-\frac{1}{9}\right) = 9 \ln 3 - 3 + \frac{1}{9}
The value of the definite integral is:
9ln⁡3−269 \boxed{9 \ln 3 - \frac{26}{9}}

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