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

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

From the diagram, the shaded region is bounded by the graph \(y=x^2-2x\), the \(x\)-axis, and the lines \(x=2\) and \(x=5\).

On the interval \([2,5]\), the function is non-negative, so

\(S=\displaystyle\int_{2}^{5}(x^2-2x)\,dx\).

An antiderivative is

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

Therefore,

\(S=\left(\dfrac{x^3}{3}-x^2\right)\Big|_{2}^{5}=\left(\dfrac{125}{3}-25\right)-\left(\dfrac83-4\right)=18\).

From the diagram, the shaded region lies between the graph \(y=1-x^3\), the \(x\)-axis, and the lines \(x=-1\) and \(x=1\).

The function is non-negative on this interval, so

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

An antiderivative is

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

Hence

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

From the diagram, the region is bounded by the graph \(y=8-x^2\), the \(x\)-axis, and the lines \(x=-2\) and \(x=1\).

Therefore, the area is

\(S=\displaystyle\int_{-2}^{1}(8-x^2)\,dx\).

An antiderivative is

\(8x-\dfrac{x^3}{3}\).

Hence

\(S=\left(8x-\dfrac{x^3}{3}\right)\Big|_{-2}^{1}=\left(8-\dfrac13\right)-\left(-16+\dfrac83\right)=21\).

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

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

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

An antiderivative is

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

Hence

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

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

Therefore,

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

An antiderivative is

\(\dfrac{x^5}{5}+2x\).

Hence

\(S=\left(\dfrac{x^5}{5}+2x\right)\Big|_{-1}^{1}=\left(\dfrac15+2\right)-\left(-\dfrac15-2\right)=\dfrac{22}{5}\).

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

The area is

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

An antiderivative is

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

Therefore,

\(S=\left(-\dfrac{x^3}{3}+2x^2\right)\Big|_{1}^{4}=\left(-\dfrac{64}{3}+32\right)-\left(-\dfrac13+2\right)=9\).

From the diagram, the shaded region is bounded by the graph \(y=-x^2-5x\), the \(x\)-axis, and the lines \(x=-4\) and \(x=-1\).

Therefore,

\(S=\displaystyle\int_{-4}^{-1}(-x^2-5x)\,dx\).

An antiderivative is

\(-\dfrac{x^3}{3}-\dfrac{5x^2}{2}\).

Hence

\(S=\left(-\dfrac{x^3}{3}-\dfrac{5x^2}{2}\right)\Big|_{-4}^{-1}=\left(\dfrac13-\dfrac52\right)-\left(\dfrac{64}{3}-40\right)=\dfrac{33}{2}\).

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

On this interval, the function is non-negative, so

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

An antiderivative is

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

Therefore,

\(S=\left(\dfrac{x^3}{3}-x\right)\Big|_{1}^{4}=\left(\dfrac{64}{3}-4\right)-\left(\dfrac13-1\right)=18\).

From the diagram, the region is bounded by the graph \(y=-x^4+3\), the \(x\)-axis, and the lines \(x=-1\) and \(x=1\).

Therefore, the area is

\(S=\displaystyle\int_{-1}^{1}(-x^4+3)\,dx\).

An antiderivative is

\(-\dfrac{x^5}{5}+3x\).

Hence

\(S=\left(-\dfrac{x^5}{5}+3x\right)\Big|_{-1}^{1}=\left(-\dfrac15+3\right)-\left(\dfrac15-3\right)=\dfrac{28}{5}\).

From the diagram, the shaded region is bounded by the graph \(y=5-x^2\), the \(x\)-axis, and the lines \(x=-2\) and \(x=1\).

So

\(S=\displaystyle\int_{-2}^{1}(5-x^2)\,dx\).

An antiderivative is

\(5x-\dfrac{x^3}{3}\).

Hence

\(S=\left(5x-\dfrac{x^3}{3}\right)\Big|_{-2}^{1}=\left(5-\dfrac13\right)-\left(-10+\dfrac83\right)=12\).

From the diagram, the region is bounded by the graph \(y=x^2+2x\), the \(x\)-axis, and the lines \(x=-5\) and \(x=-2\).

On the interval \([-5,-2]\), the function is non-negative, so

\(S=\displaystyle\int_{-5}^{-2}(x^2+2x)\,dx\).

An antiderivative is

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

Hence

\(S=\left(\dfrac{x^3}{3}+x^2\right)\Big|_{-5}^{-2}=\left(-\dfrac83+4\right)-\left(-\dfrac{125}{3}+25\right)=18\).