![\int\limits^{2}_{-4} [(-2x^2-8x+32)-(2x^2)]dx\\\\\\=\int\limits^{2}_{-4} (-4x^2-8x+32)dx\\\\\\=\bigg(-\dfrac{4}{3}x^3-\dfrac{8}{2}x^2+32x\bigg)\bigg|\limits^2_{-4}\\\\\\=\bigg(-\dfrac{4}{3}(2)^3-4(2)^2+32(2)\bigg)-\bigg(-\dfrac{4}{3}(-4)^3-4(-4)^2+32(-4)\bigg)\\\\\\=\bigg(-\dfrac{32}{3}-16+64\bigg)-\bigg(\dfrac{256}{3}-64-128\bigg)\\\\\\=\bigg(\dfrac{112}{3}\bigg)-\bigg(-\dfrac{320}{3}\bigg)\\\\\\=\dfrac{432}{3}\\\\\\=\large\boxed{144}](/tpl/images/0481/9412/00687.png)
![\int\limits^{2}_{-1} [(x+3)-(x^2+1)]dx\\\\\\=\int\limits^{2}_{-1} (-x^2+x+2)dx\\\\\\=\bigg(-\dfrac{x^3}{3}+\dfrac{x^2}{2}+2x\bigg)\bigg|\limits^2_{-1}\\\\\\=\bigg(-\dfrac{2^3}{3}+\dfrac{2^2}{2}+2(2)\bigg)-\bigg(-\dfrac{(-1)^3}{3}+\dfrac{(-1)^2}{2}+2(-1)\bigg)\\\\\\=\bigg(-\dfrac{8}{3}+2+4\bigg)-\bigg(\dfrac{1}{3}+\dfrac{1}{2}-2\bigg)\\\\\\=\bigg(\dfrac{10}{3}\bigg)-\bigg(-\dfrac{7}{6}\bigg)\\\\\\=\dfrac{9}{2}\\\\\\=\large\boxed{4.5}](/tpl/images/0481/9412/40b4c.png)
Hello there!
can i get some with this calculus practice problem? i don't think my an...
Mathematics, 29.01.2020 12:54, cilicia123345
Hello there!
can i get some with this calculus practice problem? i don't think my answer is right : )
show your work and details. a lot!
Answers: 2
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