2 find ∫₀¹(2x - 1)² dx\n3 find ∫₀¹(2x² - 3)⁴ xdx

2.

Answer:

$\frac{1}{3}$

Explanation:

Step1: Expand the integrand

$(2x - 1)^2=4x^{2}-4x + 1$

Step2: Integrate term - by - term

$\int_{0}^{1}(4x^{2}-4x + 1)dx=\left[\frac{4}{3}x^{3}-2x^{2}+x\right]_{0}^{1}$

Step3: Evaluate the definite integral

$\left(\frac{4}{3}(1)^{3}-2(1)^{2}+1\right)-\left(\frac{4}{3}(0)^{3}-2(0)^{2}+0\right)=\frac{4}{3}-2 + 1=\frac{4 - 6+3}{3}=\frac{1}{3}$

3.

Answer:

$-\frac{1}{20}$

Explanation:

Step1: Use substitution

Let $u = 2x^{2}-3$, then $du=4xdx$ and $xdx=\frac{1}{4}du$. When $x = 0$, $u=-3$; when $x = 1$, $u=-1$.

Step2: Rewrite the integral

$\int_{0}^{1}(2x^{2}-3)^{4}xdx=\frac{1}{4}\int_{-3}^{-1}u^{4}du$

Step3: Integrate $u^{4}$

$\frac{1}{4}\times\frac{u^{5}}{5}\big|_{-3}^{-1}=\frac{1}{20}[(-1)^{5}-(-3)^{5}]$

Step4: Evaluate the definite integral

$\frac{1}{20}(-1 + 243)=\frac{1}{20}\times(-244)=-\frac{1}{20}$