Computing Definite Integrals — Question 1

Question 1 - Solution

Combine the terms: $$\int_{0}^{1} e^{x}(x^2-2x)\,dx.$$

Use integration by parts. Let $$u=x^2-2x, \qquad dv=e^{x}\,dx.$$ Then $$du=(2x-2)\,dx, \qquad v=e^{x}.$$

Apply integration by parts: $$\int e^{x}(x^2-2x)\,dx = (x^2-2x)e^{x} - \int e^{x}(2x-2)\,dx.$$

Apply integration by parts again to the remaining integral. Let $$u=2x-2, \qquad dv=e^{x}\,dx.$$ Then $$du=2\,dx, \qquad v=e^{x}.$$

Thus, $$\int e^{x}(2x-2)\,dx = (2x-2)e^{x}-\int 2e^{x}\,dx = (2x-2)e^{x}-2e^{x}.$$

Substitute back: $$\int e^{x}(x^2-2x)\,dx = (x^2-2x)e^{x} - \bigl[(2x-2)e^{x}-2e^{x}\bigr].$$

Simplify: $$(x^2-4x+4)e^{x}.$$

Now evaluate from $0$ to $1$: $$\int_{0}^{1} e^{x}(x^2-2x)\,dx = \bigl[(x^2-4x+4)e^{x}\bigr]_{0}^{1}.$$

Compute: $$(1-4+4)e-(0-0+4) = e-4.$$

$$\boxed{e-4}$$