Question

In Exercises 55 and \(56,\) find \(K\) so that $$\int_{a}^{x} f(t) d t+K=\int_{b}^{x} f(t) d t$$ $$f(x)=\sin ^{2} x ; \quad a=0 ; \quad b=2$$

Short answer

To find the solution to this exercise, compute the integrals first, then make them equal and solve the equation for K. This will give the correct value of K that makes the two integrals equal.

Step-by-step solution

Compute the integrals

To find the stated integral values, use the integral formula \(\int_{a}^{x} f(t) d t\). First compute \(\int_{0}^{x} \sin^{2}t dt\), and then compute \(\int_{2}^{x} \sin^{2}t dt\). These integrals represent the accumulated area under the curve of the function from 0 to x, and from 2 to x, respectively.

Solve for K

After finding the values of the integrals, the equation that needs to be solved for K is \(\int_{0}^{x} \sin^{2}t dt + K = \int_{2}^{x} \sin^{2}t dt\). To solve this equation, remove the integral on both sides that are equal, simplify and isolate K.

Interpret the result

The solution to this problem is the value of K, which makes the two integrals equal when added to the integral from 0 to x. This implies that the additional K-area, when added to the first integral (0 to x), gives the same area as the second integral (2 to x).