IB Mathematics Applications & Interpretation Question Type 3: Calculating the Area Between a Function and the Y-axis Exercises

Calculate the definite integral $${\int }_{2.5}^{7.5}(x^2-5x)dx$$

Evaluate the total (unsigned) area between the curve $y=x^2-5x$ and the $x$-axis from $x=2.5$ to $x=7.5$ by integrating $|x^2-5x|$.

Evaluate the area of the region above the $x$-axis bounded by the curve $y=x^2-5x$ and the lines $x=5$ and $x=7.5$.

Compute the area of the region where the curve $y=x^2-5x$ lies below the $x$-axis, between $x=2.5$ and $x=5$.

Find the area enclosed by the curve $y=x^2-5x$, the $x$-axis, and the vertical lines $x=2.5$ and $x=7.5$.

Find the value of $c\in [2.5,7.5]$ such that the area between the curve $y=x^2-5x$ and the $x$-axis from $x=2.5$ to $x=c$ is equal to the area from $x=c$ to $x=7.5$.

Find the $x$-coordinates where the curve $y=x^2-5x$ meets the $x$-axis. Hence, calculate the total area bounded by the curve and the $x$-axis between $x=0$ and $x=7.5$.

Determine the point $c\in [2.5,7.5]$ guaranteed by the Mean Value Theorem for Integrals so that $${\int }_{2.5}^{7.5}(x^2-5x)dx=(7.5-2.5)(c^2-5c).$$

Find the average value of $f(x)=x^2-5x$ on the interval $[2.5,7.5]$ and verify that $\text{area}=f_{\text{avg}}\times (7.5-2.5)$, where $\text{area}$ represents the definite integral of the function over the given interval.

Use the linearity of the integral to show $${\int }_{2.5}^{7.5}(x^2-5x)dx={\int }_{2.5}^{7.5}{x}^{2}dx-5{\int }_{2.5}^{7.5}xdx$$ and then compute its value.

Use Simpson’s rule with $n=4$ to approximate the value of ${\int }_{2.5}^{7.5}(x^2-5x)dx$.