Questions: Given ∫ from 0 to b of sin^3(x)cos(x)dx=0.15 and ∫ from - to 0 of sin^3(x)cos(x)dx=-0.2, evaluate: a) ∫ from -a to b of sin^3(x)cos(x)dx= b) ∫ from b to 0 of sin^3(x)cos(x)dx= c) ∫ from -a to -a of 4sin^3(x)cos(x)dx= d) ∫ from -a to a of 4sin^3(x)cos(x)dx=

Solution Steps

To solve the given integrals, we can use the properties of definite integrals and the given values.

a) Use the property of definite integrals that states the integral from \(-a\) to \(b\) can be split into the sum of integrals from \(-a\) to \(0\) and from \(0\) to \(b\).

b) Use the property of definite integrals that allows reversing the limits of integration, which changes the sign of the integral.

c) Recognize that the integral from \(-a\) to \(-a\) is zero because the limits of integration are the same.

To find \(\int_{-a}^{b} \sin^3(x) \cos(x) \, dx\), we use the property of definite integrals that allows us to split the integral into two parts:

\[ \int_{-a}^{b} \sin^3(x) \cos(x) \, dx = \int_{-a}^{0} \sin^3(x) \cos(x) \, dx + \int_{0}^{b} \sin^3(x) \cos(x) \, dx \]

Given that \(\int_{-a}^{0} \sin^3(x) \cos(x) \, dx = -0.2\) and \(\int_{0}^{b} \sin^3(x) \cos(x) \, dx = 0.15\), we have:

\[ \int_{-a}^{b} \sin^3(x) \cos(x) \, dx = -0.2 + 0.15 = -0.05 \]

To find \(\int_{b}^{0} \sin^3(x) \cos(x) \, dx\), we use the property of definite integrals that states reversing the limits of integration changes the sign of the integral:

\[ \int_{b}^{0} \sin^3(x) \cos(x) \, dx = -\int_{0}^{b} \sin^3(x) \cos(x) \, dx \]

Given that \(\int_{0}^{b} \sin^3(x) \cos(x) \, dx = 0.15\), we have:

\[ \int_{b}^{0} \sin^3(x) \cos(x) \, dx = -0.15 \]

The integral from \(-a\) to \(-a\) is zero because the limits of integration are the same:

\[ \int_{-a}^{-a} 4 \sin^3(x) \cos(x) \, dx = 0 \]

Final Answer