Solve ∫ (from 1 to frac{ 3) of {2}}left(x^2-3x+2right)^2 wrt x

$$\int_{ 1 }^{ \frac{ 3 }{ 2 } } { \left( { x }^{ 2 } -3x+2 \right) }^{ 2 } d x$$

Evaluate

$$\frac{1}{60}\approx 0.016666667$$

Solution Steps

Square x^{2}-3x+2.

$$\int _{1}^{\frac{3}{2}}x^{4}-6x^{3}+13x^{2}-12x+4\mathrm{d}x$$

Evaluate the indefinite integral first.

$$\int x^{4}-6x^{3}+13x^{2}-12x+4\mathrm{d}x$$

Integrate the sum term by term.

$$\int x^{4}\mathrm{d}x+\int -6x^{3}\mathrm{d}x+\int 13x^{2}\mathrm{d}x+\int -12x\mathrm{d}x+\int 4\mathrm{d}x$$

Factor out the constant in each of the terms.

$$\int x^{4}\mathrm{d}x-6\int x^{3}\mathrm{d}x+13\int x^{2}\mathrm{d}x-12\int x\mathrm{d}x+\int 4\mathrm{d}x$$

Since \int x^{k}\mathrm{d}x=\frac{x^{k+1}}{k+1} for k\neq -1, replace \int x^{4}\mathrm{d}x with \frac{x^{5}}{5}.

$$\frac{x^{5}}{5}-6\int x^{3}\mathrm{d}x+13\int x^{2}\mathrm{d}x-12\int x\mathrm{d}x+\int 4\mathrm{d}x$$

Since \int x^{k}\mathrm{d}x=\frac{x^{k+1}}{k+1} for k\neq -1, replace \int x^{3}\mathrm{d}x with \frac{x^{4}}{4}. Multiply -6 times \frac{x^{4}}{4}.

$$\frac{x^{5}}{5}-\frac{3x^{4}}{2}+13\int x^{2}\mathrm{d}x-12\int x\mathrm{d}x+\int 4\mathrm{d}x$$

Since \int x^{k}\mathrm{d}x=\frac{x^{k+1}}{k+1} for k\neq -1, replace \int x^{2}\mathrm{d}x with \frac{x^{3}}{3}. Multiply 13 times \frac{x^{3}}{3}.

$$\frac{x^{5}}{5}-\frac{3x^{4}}{2}+\frac{13x^{3}}{3}-12\int x\mathrm{d}x+\int 4\mathrm{d}x$$

Since \int x^{k}\mathrm{d}x=\frac{x^{k+1}}{k+1} for k\neq -1, replace \int x\mathrm{d}x with \frac{x^{2}}{2}. Multiply -12 times \frac{x^{2}}{2}.

$$\frac{x^{5}}{5}-\frac{3x^{4}}{2}+\frac{13x^{3}}{3}-6x^{2}+\int 4\mathrm{d}x$$

Find the integral of 4 using the table of common integrals rule \int a\mathrm{d}x=ax.

$$\frac{x^{5}}{5}-\frac{3x^{4}}{2}+\frac{13x^{3}}{3}-6x^{2}+4x$$

The definite integral is the antiderivative of the expression evaluated at the upper limit of integration minus the antiderivative evaluated at the lower limit of integration.

$$\frac{\left(\frac{3}{2}\right)^{5}}{5}-\frac{3}{2}\times \left(\frac{3}{2}\right)^{4}+\frac{13}{3}\times \left(\frac{3}{2}\right)^{3}-6\times \left(\frac{3}{2}\right)^{2}+4\times \frac{3}{2}-\left(\frac{1^{5}}{5}-\frac{3}{2}\times 1^{4}+\frac{13}{3}\times 1^{3}-6\times 1^{2}+4\times 1\right)$$

Simplify.

$$\frac{1}{60}$$