Question

fundamental theorem of calculus: problem 1
(1 point)
evaluate the definite integral
$$\int_{3}^{7}(4 - x^{2})dx$$

Explanation:

Step1: Expand the integrand

$$(4 - x)^{2}=16-8x + x^{2}$$

Step2: Integrate term - by - term

$$\int_{-3}^{2}(16-8x + x^{2})dx=\int_{-3}^{2}16dx-\int_{-3}^{2}8xdx+\int_{-3}^{2}x^{2}dx$$
Using the power rule for integration \(\int x^{n}dx=\frac{x^{n + 1}}{n+1}+C(n
eq - 1)\) and \(\int kdx=kx + C\) (where \(k\) is a constant):

• \(\int_{-3}^{2}16dx=16x\big|_{-3}^{2}=16\times(2)-16\times(-3)=32 + 48 = 80\)
• \(\int_{-3}^{2}8xdx=8\times\frac{x^{2}}{2}\big|_{-3}^{2}=4x^{2}\big|_{-3}^{2}=4\times(2^{2})-4\times((-3)^{2})=16 - 36=-20\)
• \(\int_{-3}^{2}x^{2}dx=\frac{x^{3}}{3}\big|_{-3}^{2}=\frac{2^{3}}{3}-\frac{(-3)^{3}}{3}=\frac{8}{3}+\frac{27}{3}=\frac{35}{3}\)

Step3: Combine the results

$$\int_{-3}^{2}(16-8x + x^{2})dx=80-(-20)+\frac{35}{3}=100+\frac{35}{3}=\frac{300 + 35}{3}=\frac{335}{3}$$

Answer:

\(\frac{335}{3}\)