From the diagram, the area is bounded by the graph \(y=x^3+1\), the \(x\)-axis, and the vertical lines \(x=-1\) and \(x=1\).

Since the function is non-negative on this interval, the area is equal to the definite integral:

\(S=\displaystyle\int_{-1}^{1}(x^3+1)\,dx\).

Find an antiderivative:

\(\int (x^3+1)\,dx=\dfrac{x^4}{4}+x\).

Hence

\(S=\left(\dfrac{x^4}{4}+x\right)\Big|_{-1}^{1}=\left(\dfrac14+1\right)-\left(\dfrac14-1\right)=2\).

From the diagram, the shaded region lies between the graph \(y=6-x^2\), the \(x\)-axis, and the lines \(x=-1\) and \(x=2\).

Therefore, its area is

\(S=\displaystyle\int_{-1}^{2}(6-x^2)\,dx\).

An antiderivative of \(6-x^2\) is

\(6x-\dfrac{x^3}{3}\).

Hence

\(S=\left(6x-\dfrac{x^3}{3}\right)\Big|_{-1}^{2}=\left(12-\dfrac{8}{3}\right)-\left(-6+\dfrac13\right)=15\).

From the diagram, the shaded region is bounded by the graph \(y=x^2-x\), the \(x\)-axis, and the lines \(x=1\) and \(x=4\).

On the interval \([1,4]\), the function is non-negative, so

\(S=\displaystyle\int_{1}^{4}(x^2-x)\,dx\).

An antiderivative is

\(\dfrac{x^3}{3}-\dfrac{x^2}{2}\).

Hence

\(S=\left(\dfrac{x^3}{3}-\dfrac{x^2}{2}\right)\Big|_{1}^{4}=\left(\dfrac{64}{3}-8\right)-\left(\dfrac13-\dfrac12\right)=\dfrac{40}{3}+\dfrac16=\dfrac{81}{6}=\dfrac{27}{2}\).

From the diagram, the region is bounded by the graph \(y=-x^3+2\), the \(x\)-axis, and the lines \(x=-1\) and \(x=1\).

On this interval, the function is positive, so

\(S=\displaystyle\int_{-1}^{1}(-x^3+2)\,dx\).

An antiderivative is

\(-\dfrac{x^4}{4}+2x\).

Therefore,

\(S=\left(-\dfrac{x^4}{4}+2x\right)\Big|_{-1}^{1}=\left(-\dfrac14+2\right)-\left(-\dfrac14-2\right)=4\).

From the diagram, the region is bounded by the graph \(y=7-x^2\), the \(x\)-axis, and the lines \(x=-2\) and \(x=1\).

Therefore,

\(S=\displaystyle\int_{-2}^{1}(7-x^2)\,dx\).

An antiderivative is

\(7x-\dfrac{x^3}{3}\).

Hence

\(S=\left(7x-\dfrac{x^3}{3}\right)\Big|_{-2}^{1}=\left(7-\dfrac13\right)-\left(-14+\dfrac83\right)=18\).