If the area bounded by y=||x|2 - 4 |x + 3|| and the x -axis from x=1 to x=3 is (p/q) square units (where, p & q are coprime), then the value of p+q is

Question

If the area bounded by y=||x|2 - 4 |x + 3|| and the x -axis from x=1 to x=3 is (p/q) square units (where, p & q are coprime), then the value of p+q is (A)  (B)  (C)  (D)

Tardigrade Admin · Accepted Answer

Area =| displaystyle ∫ 13 (4 x - x2 + 12) d x| =(94/3) square units. <img class="img-fluid question-image" alt="Solution" src="https://cdn.tardigrade.in/q/nta/m-10jozv8xsykyurpx.jpg" />

Q. If the area bounded by $y = ∣ ∣ ∣ x ∣^{2} - 4 ∣ x + 3 ∣ ∣ ∣$ and the $x$ -axis from $x = 1$ to $x = 3$ is $\frac{p}{q}$ square units (where, $p$ & $q$ are coprime), then the value of $p + q$ is

Area $= ∣ ∣ \int_{1}^{3} (4 x - x^{2} + 12) d x ∣ ∣$ $= \frac{94}{3}$ square units.

Q. If the area bounded by y=∣∣​∣x∣2−4∣x+3∣∣∣​ and the x -axis from x=1 to x=3 is qp​ square units (where, p & q are coprime), then the value of p+q is

Area =∣∣​∫13​(4x−x2+12)dx∣∣​ =394​ square units.

Q. If the area bounded by $y = ∣ ∣ ∣ x ∣^{2} - 4 ∣ x + 3 ∣ ∣ ∣$ and the $x$ -axis from $x = 1$ to $x = 3$ is $\frac{p}{q}$ square units (where, $p$ & $q$ are coprime), then the value of $p + q$ is

Area $= ∣ ∣ \int_{1}^{3} (4 x - x^{2} + 12) d x ∣ ∣$ $= \frac{94}{3}$ square units.