Answer to Question #277778 in Calculus for n hasan

Question

a,Calculate the area under the curve 𝑦 = 𝑥3 + 4𝑥 + 1 from x=-3 to x=3. 5

b) Evaluate: ∫4 𝑥𝑒𝑥 dx.

Answer 1

a)

$\displaystyle\int\limits_{-3}^{3,5} (x^3 + 4x + 1)dx = (\dfrac{x^4}{4} + 2x^2 + x)\big|_{-3}^{3,5} = \Big(\dfrac{(3,5)^4}{4} - \dfrac{(-3)^4}{4} \Big) + \\ + \Big( 2*(3,5)^2 - 2 * (-3)^{2} \Big) + (3,5 - (-3)) = \dfrac{150,0625 - 81}{4} + 2*(12, 25 - 9) + 6,5 = \\ = 17,265625 + 6,5 + 6,5 = 30, 265625$

b) $\int 4xe^{x} dx = 4 \int xe^{x} dx$

$u = x \\ v' = e^{x} \\ u' = 1 \\ v = e^{x}$

$uv' = (uv)' - u'v$

$4 \int xe^{x} dx = 4(xe^x - \int (1*e^x) dx) = 4(xe^x - e^x)+c$