Answer to Question #87994 in Calculus for Joshua

Question

5.Evalute ∫x^2(3−10x^3)dx
a.
1
--- (3−10x^3)^5)+c
150
b.
1
--- (1−10x^2)^5)+c
10
c.
1
--- (3−20x^3)^5)+c
15
d.
1
--- (3−2x^3)^5)+c
100

6.Evaluate ∫3e^x+5cos(x)−10sec^2(x)dx
a.3e^2+5sinx−10tanx+c
b.3e^x+cosx−10tanx+c
c.3ex^+5sinx−10secx+c
d.2e^x−x−10tanx+c

Answer 1

Task #5.

$\int x^2(3-10x^3)dx=\int3x^2-10x^5dx=x^3-(10/6)*x^6+c$

where c is constant.

Task #6.

$\int 3e^x+5cosx-10sec^2xdx = 3e^x+5sinx-10tanx +c$

where c is constant.